Develop Reference

How to get more than one position using index()? [duplicate]

index() will give the first occurrence of an item in a list. Is there a neat trick which returns all indices in a list for an element?

You can use a list comprehension with enumerate:
indices = [i for i, x in enumerate(my_list) if x == "whatever"]
The iterator enumerate(my_list) yields pairs (index, item) for each item in the list. Using i, x as loop variable target unpacks these pairs into the index i and the list item x. We filter down to all x that match our criterion, and select the indices i of these elements.

While not a solution for lists directly, numpy really shines for this sort of thing:
import numpy as np
values = np.array([1,2,3,1,2,4,5,6,3,2,1])
searchval = 3
ii = np.where(values == searchval)[0]
returns:
ii ==>array([2, 8])
This can be significantly faster for lists (arrays) with a large number of elements vs some of the other solutions.

A solution using list.index:
def indices(lst, element):
result = []
offset = -1
while True:
try:
offset = lst.index(element, offset+1)
except ValueError:
return result
result.append(offset)
It's much faster than the list comprehension with enumerate, for large lists. It is also much slower than the numpy solution if you already have the array, otherwise the cost of converting outweighs the speed gain (tested on integer lists with 100, 1000 and 10000 elements).
NOTE: A note of caution based on Chris_Rands' comment: this solution is faster than the list comprehension if the results are sufficiently sparse, but if the list has many instances of the element that is being searched (more than ~15% of the list, on a test with a list of 1000 integers), the list comprehension is faster.

How about:
In [1]: l=[1,2,3,4,3,2,5,6,7]
In [2]: [i for i,val in enumerate(l) if val==3]
Out[2]: [2, 4]

more_itertools.locate finds indices for all items that satisfy a condition.
from more_itertools import locate
list(locate([0, 1, 1, 0, 1, 0, 0]))
# [1, 2, 4]
list(locate(['a', 'b', 'c', 'b'], lambda x: x == 'b'))
# [1, 3]
more_itertools is a third-party library > pip install more_itertools.

occurrences = lambda s, lst: (i for i,e in enumerate(lst) if e == s)
list(occurrences(1, [1,2,3,1])) # = [0, 3]

Or Use range (python 3):
l=[i for i in range(len(lst)) if lst[i]=='something...']
For (python 2):
l=[i for i in xrange(len(lst)) if lst[i]=='something...']
And then (both cases):
print(l)
Is as expected.

There’s an answer using np.where to find the indices of a single value, which is not faster than a list-comprehension, if the time to convert a list to an array is included
The overhead of importing numpy and converting a list to a numpy.array probably makes using numpy a less efficient option for most circumstances. A careful timing analysis would be necessary.
In cases where multiple functions/operations will need to be performed on the list, converting the list to an array, and then using numpy functions will likely be a faster option.
This solution uses np.where and np.unique to find the indices of all unique elements in a list.
Using np.where on an array (including the time to convert the list to an array) is slightly slower than a list-comprehension on a list, for finding all indices of all unique elements.
This has been tested on an 2M element list with 4 unique values, and the size of the list/array and number of unique elements will have an impact.
Other solutions using numpy on an array can be found in Get a list of all indices of repeated elements in a numpy array
Tested in [python 3.10.4, numpy 1.23.1] and [python 3.11.0, numpy 1.23.4]
import numpy as np
import random # to create test list
# create sample list
random.seed(365)
l = [random.choice(['s1', 's2', 's3', 's4']) for _ in range(20)]
# convert the list to an array for use with these numpy methods
a = np.array(l)
# create a dict of each unique entry and the associated indices
idx = {v: np.where(a == v)[0].tolist() for v in np.unique(a)}
# print(idx)
{'s1': [7, 9, 10, 11, 17],
's2': [1, 3, 6, 8, 14, 18, 19],
's3': [0, 2, 13, 16],
's4': [4, 5, 12, 15]}
%timeit on a 2M element list with 4 unique str elements
# create 2M element list
random.seed(365)
l = [random.choice(['s1', 's2', 's3', 's4']) for _ in range(2000000)]
Functions
def test1():
# np.where: convert list to array and find indices of a single element
a = np.array(l)
return np.where(a == 's1')
def test2():
# list-comprehension: on list l and find indices of a single element
return [i for i, x in enumerate(l) if x == "s1"]
def test3():
# filter: on list l and find indices of a single element
return list(filter(lambda i: l[i]=="s1", range(len(l))))
def test4():
# use np.where and np.unique to find indices of all unique elements: convert list to array
a = np.array(l)
return {v: np.where(a == v)[0].tolist() for v in np.unique(a)}
def test5():
# list comprehension inside dict comprehension: on list l and find indices of all unique elements
return {req_word: [idx for idx, word in enumerate(l) if word == req_word] for req_word in set(l)}
Function Call
%timeit test1()
%timeit test2()
%timeit test3()
%timeit test4()
%timeit test5()
Result python 3.10.4
214 ms ± 19.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
85.1 ms ± 1.48 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
146 ms ± 1.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
365 ms ± 11.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
360 ms ± 5.82 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Result python 3.11.0
209 ms ± 15.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
70.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
132 ms ± 4.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
371 ms ± 20.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
314 ms ± 15.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Getting all the occurrences and the position of one or more (identical) items in a list
With enumerate(alist) you can store the first element (n) that is the index of the list when the element x is equal to what you look for.
>>> alist = ['foo', 'spam', 'egg', 'foo']
>>> foo_indexes = [n for n,x in enumerate(alist) if x=='foo']
>>> foo_indexes
[0, 3]
>>>
Let's make our function findindex
This function takes the item and the list as arguments and return the position of the item in the list, like we saw before.
def indexlist(item2find, list_or_string):
"Returns all indexes of an item in a list or a string"
return [n for n,item in enumerate(list_or_string) if item==item2find]
print(indexlist("1", "010101010"))
Output
[1, 3, 5, 7]
Simple
for n, i in enumerate([1, 2, 3, 4, 1]):
if i == 1:
print(n)
Output:
0
4

One more solution(sorry if duplicates) for all occurrences:
values = [1,2,3,1,2,4,5,6,3,2,1]
map(lambda val: (val, [i for i in xrange(len(values)) if values[i] == val]), values)

Using filter() in python2.
>>> q = ['Yeehaw', 'Yeehaw', 'Googol', 'B9', 'Googol', 'NSM', 'B9', 'NSM', 'Dont Ask', 'Googol']
>>> filter(lambda i: q[i]=="Googol", range(len(q)))
[2, 4, 9]

If you need to search for all element's positions between certain indices, you can state them:
[i for i,x in enumerate([1,2,3,2]) if x==2 & 2<= i <=3] # -> [3]

A dynamic list comprehension based solution incase we do not know in advance which element:
lst = ['to', 'be', 'or', 'not', 'to', 'be']
{req_word: [idx for idx, word in enumerate(lst) if word == req_word] for req_word in set(lst)}
results in:
{'be': [1, 5], 'or': [2], 'to': [0, 4], 'not': [3]}
You can think of all other ways along the same lines as well but with index() you can find only one index although you can set occurrence number yourself.

Using a for-loop:
Answers with enumerate and a list comprehension are more pythonic, but not necessarily faster. However, this answer is aimed at students who may not be allowed to use some of those built-in functions.
create an empty list, indices
create the loop with for i in range(len(x)):, which essentially iterates through a list of index locations [0, 1, 2, 3, ..., len(x)-1]
in the loop, add any i, where x[i] is a match to value, to indices
x[i] accesses the list by index
def get_indices(x: list, value: int) -> list:
indices = list()
for i in range(len(x)):
if x[i] == value:
indices.append(i)
return indices
n = [1, 2, 3, -50, -60, 0, 6, 9, -60, -60]
print(get_indices(n, -60))
>>> [4, 8, 9]
The functions, get_indices, are implemented with type hints. In this case, the list, n, is a bunch of ints, therefore we search for value, also defined as an int.
Using a while-loop and .index:
With .index, use try-except for error handling, because a ValueError will occur if value is not in the list.
def get_indices(x: list, value: int) -> list:
indices = list()
i = 0
while True:
try:
# find an occurrence of value and update i to that index
i = x.index(value, i)
# add i to the list
indices.append(i)
# advance i by 1
i += 1
except ValueError as e:
break
return indices
print(get_indices(n, -60))
>>> [4, 8, 9]

You can create a defaultdict
from collections import defaultdict
d1 = defaultdict(int) # defaults to 0 values for keys
unq = set(lst1) # lst1 = [1, 2, 2, 3, 4, 1, 2, 7]
for each in unq:
d1[each] = lst1.count(each)
else:
print(d1)

If you are using Python 2, you can achieve the same functionality with this:
f = lambda my_list, value:filter(lambda x: my_list[x] == value, range(len(my_list)))
Where my_list is the list you want to get the indexes of, and value is the value searched. Usage:
f(some_list, some_element)

Create a generator
Generators are fast and use a tiny memory footprint. They give you flexibility in how you use the result.
def indices(iter, val):
"""Generator: Returns all indices of val in iter
Raises a ValueError if no val does not occur in iter
Passes on the AttributeError if iter does not have an index method (e.g. is a set)
"""
i = -1
NotFound = False
while not NotFound:
try:
i = iter.index(val, i+1)
except ValueError:
NotFound = True
else:
yield i
if i == -1:
raise ValueError("No occurrences of {v} in {i}".format(v = val, i = iter))
The above code can be use to create a list of the indices: list(indices(input,value)); use them as dictionary keys: dict(indices(input,value)); sum them: sum(indices(input,value)); in a for loop for index_ in indices(input,value):; ...etc... without creating an interim list/tuple or similar.
In a for loop you will get your next index back when you call for it, without waiting for all the others to be calculated first. That means: if you break out of the loop for some reason you save the time needed to find indices you never needed.
How it works
Call .index on the input iter to find the next occurrence of
val
Use the second parameter to .index to start at the point
after the last found occurrence
Yield the index
Repeat until index raises a ValueError
Alternative versions
I tried four different versions for flow control; two EAFP (using try - except) and two TBYL (with a logical test in the while statement):
"WhileTrueBreak": while True: ... except ValueError: break. Surprisingly, this was usually a touch slower than option 2 and (IMV) less readable
"WhileErrFalse": Using a bool variable err to identify when a ValueError is raised. This is generally the fastest and more readable than 1
"RemainingSlice": Check whether val is in the remaining part of the input using slicing: while val in iter[i:]. Unsurprisingly, this does not scale well
"LastOccurrence": Check first where the last occurrence is, keep going while i < last
The overall performance differences between 1,2 and 4 are negligible, so it comes down to personal style and preference. Given that .index uses ValueError to let you know it didn't find anything, rather than e.g. returning None, an EAFP-approach seems fitting to me.
Here are the 4 code variants and results from timeit (in milliseconds) for different lengths of input and sparsity of matches
#version("WhileTrueBreak", versions)
def indices2(iter, val):
i = -1
while True:
try:
i = iter.index(val, i+1)
except ValueError:
break
else:
yield i
#version("WhileErrFalse", versions)
def indices5(iter, val):
i = -1
err = False
while not err:
try:
i = iter.index(val, i+1)
except ValueError:
err = True
else:
yield i
#version("RemainingSlice", versions)
def indices1(iter, val):
i = 0
while val in iter[i:]:
i = iter.index(val, i)
yield i
i += 1
#version("LastOccurrence", versions)
def indices4(iter,val):
i = 0
last = len(iter) - tuple(reversed(iter)).index(val)
while i < last:
i = iter.index(val, i)
yield i
i += 1
Length: 100, Ocurrences: 4.0%
{'WhileTrueBreak': 0.0074799987487494946, 'WhileErrFalse': 0.006440002471208572, 'RemainingSlice': 0.01221001148223877, 'LastOccurrence': 0.00801000278443098}
Length: 1000, Ocurrences: 1.2%
{'WhileTrueBreak': 0.03101000329479575, 'WhileErrFalse': 0.0278000021353364, 'RemainingSlice': 0.08278000168502331, 'LastOccurrence': 0.03986000083386898}
Length: 10000, Ocurrences: 2.05%
{'WhileTrueBreak': 0.18062000162899494, 'WhileErrFalse': 0.1810499932616949, 'RemainingSlice': 2.9145700042136014, 'LastOccurrence': 0.2049500006251037}
Length: 100000, Ocurrences: 1.977%
{'WhileTrueBreak': 1.9361200043931603, 'WhileErrFalse': 1.7280600033700466, 'RemainingSlice': 254.4725100044161, 'LastOccurrence': 1.9101499929092824}
Length: 100000, Ocurrences: 9.873%
{'WhileTrueBreak': 2.832529996521771, 'WhileErrFalse': 2.9984100023284554, 'RemainingSlice': 1132.4922299943864, 'LastOccurrence': 2.6660699979402125}
Length: 100000, Ocurrences: 25.058%
{'WhileTrueBreak': 5.119729996658862, 'WhileErrFalse': 5.2082200068980455, 'RemainingSlice': 2443.0577100021765, 'LastOccurrence': 4.75954000139609}
Length: 100000, Ocurrences: 49.698%
{'WhileTrueBreak': 9.372120001353323, 'WhileErrFalse': 8.447749994229525, 'RemainingSlice': 5042.717969999649, 'LastOccurrence': 8.050809998530895}

Here is a time performance comparison between using np.where vs list_comprehension. Seems like np.where is faster on average.
# np.where
start_times = []
end_times = []
for i in range(10000):
start = time.time()
start_times.append(start)
temp_list = np.array([1,2,3,3,5])
ixs = np.where(temp_list==3)[0].tolist()
end = time.time()
end_times.append(end)
print("Took on average {} seconds".format(
np.mean(end_times)-np.mean(start_times)))
Took on average 3.81469726562e-06 seconds
# list_comprehension
start_times = []
end_times = []
for i in range(10000):
start = time.time()
start_times.append(start)
temp_list = np.array([1,2,3,3,5])
ixs = [i for i in range(len(temp_list)) if temp_list[i]==3]
end = time.time()
end_times.append(end)
print("Took on average {} seconds".format(
np.mean(end_times)-np.mean(start_times)))
Took on average 4.05311584473e-06 seconds

Make a list of values from another list that meet a condition [duplicate]

index() will give the first occurrence of an item in a list. Is there a neat trick which returns all indices in a list for an element?

You can use a list comprehension with enumerate:
indices = [i for i, x in enumerate(my_list) if x == "whatever"]
The iterator enumerate(my_list) yields pairs (index, item) for each item in the list. Using i, x as loop variable target unpacks these pairs into the index i and the list item x. We filter down to all x that match our criterion, and select the indices i of these elements.

While not a solution for lists directly, numpy really shines for this sort of thing:
import numpy as np
values = np.array([1,2,3,1,2,4,5,6,3,2,1])
searchval = 3
ii = np.where(values == searchval)[0]
returns:
ii ==>array([2, 8])
This can be significantly faster for lists (arrays) with a large number of elements vs some of the other solutions.

A solution using list.index:
def indices(lst, element):
result = []
offset = -1
while True:
try:
offset = lst.index(element, offset+1)
except ValueError:
return result
result.append(offset)
It's much faster than the list comprehension with enumerate, for large lists. It is also much slower than the numpy solution if you already have the array, otherwise the cost of converting outweighs the speed gain (tested on integer lists with 100, 1000 and 10000 elements).
NOTE: A note of caution based on Chris_Rands' comment: this solution is faster than the list comprehension if the results are sufficiently sparse, but if the list has many instances of the element that is being searched (more than ~15% of the list, on a test with a list of 1000 integers), the list comprehension is faster.

How about:
In [1]: l=[1,2,3,4,3,2,5,6,7]
In [2]: [i for i,val in enumerate(l) if val==3]
Out[2]: [2, 4]

more_itertools.locate finds indices for all items that satisfy a condition.
from more_itertools import locate
list(locate([0, 1, 1, 0, 1, 0, 0]))
# [1, 2, 4]
list(locate(['a', 'b', 'c', 'b'], lambda x: x == 'b'))
# [1, 3]
more_itertools is a third-party library > pip install more_itertools.

occurrences = lambda s, lst: (i for i,e in enumerate(lst) if e == s)
list(occurrences(1, [1,2,3,1])) # = [0, 3]

Or Use range (python 3):
l=[i for i in range(len(lst)) if lst[i]=='something...']
For (python 2):
l=[i for i in xrange(len(lst)) if lst[i]=='something...']
And then (both cases):
print(l)
Is as expected.

There’s an answer using np.where to find the indices of a single value, which is not faster than a list-comprehension, if the time to convert a list to an array is included
The overhead of importing numpy and converting a list to a numpy.array probably makes using numpy a less efficient option for most circumstances. A careful timing analysis would be necessary.
In cases where multiple functions/operations will need to be performed on the list, converting the list to an array, and then using numpy functions will likely be a faster option.
This solution uses np.where and np.unique to find the indices of all unique elements in a list.
Using np.where on an array (including the time to convert the list to an array) is slightly slower than a list-comprehension on a list, for finding all indices of all unique elements.
This has been tested on an 2M element list with 4 unique values, and the size of the list/array and number of unique elements will have an impact.
Other solutions using numpy on an array can be found in Get a list of all indices of repeated elements in a numpy array
Tested in [python 3.10.4, numpy 1.23.1] and [python 3.11.0, numpy 1.23.4]
import numpy as np
import random # to create test list
# create sample list
random.seed(365)
l = [random.choice(['s1', 's2', 's3', 's4']) for _ in range(20)]
# convert the list to an array for use with these numpy methods
a = np.array(l)
# create a dict of each unique entry and the associated indices
idx = {v: np.where(a == v)[0].tolist() for v in np.unique(a)}
# print(idx)
{'s1': [7, 9, 10, 11, 17],
's2': [1, 3, 6, 8, 14, 18, 19],
's3': [0, 2, 13, 16],
's4': [4, 5, 12, 15]}
%timeit on a 2M element list with 4 unique str elements
# create 2M element list
random.seed(365)
l = [random.choice(['s1', 's2', 's3', 's4']) for _ in range(2000000)]
Functions
def test1():
# np.where: convert list to array and find indices of a single element
a = np.array(l)
return np.where(a == 's1')
def test2():
# list-comprehension: on list l and find indices of a single element
return [i for i, x in enumerate(l) if x == "s1"]
def test3():
# filter: on list l and find indices of a single element
return list(filter(lambda i: l[i]=="s1", range(len(l))))
def test4():
# use np.where and np.unique to find indices of all unique elements: convert list to array
a = np.array(l)
return {v: np.where(a == v)[0].tolist() for v in np.unique(a)}
def test5():
# list comprehension inside dict comprehension: on list l and find indices of all unique elements
return {req_word: [idx for idx, word in enumerate(l) if word == req_word] for req_word in set(l)}
Function Call
%timeit test1()
%timeit test2()
%timeit test3()
%timeit test4()
%timeit test5()
Result python 3.10.4
214 ms ± 19.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
85.1 ms ± 1.48 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
146 ms ± 1.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
365 ms ± 11.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
360 ms ± 5.82 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Result python 3.11.0
209 ms ± 15.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
70.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
132 ms ± 4.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
371 ms ± 20.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
314 ms ± 15.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Getting all the occurrences and the position of one or more (identical) items in a list
With enumerate(alist) you can store the first element (n) that is the index of the list when the element x is equal to what you look for.
>>> alist = ['foo', 'spam', 'egg', 'foo']
>>> foo_indexes = [n for n,x in enumerate(alist) if x=='foo']
>>> foo_indexes
[0, 3]
>>>
Let's make our function findindex
This function takes the item and the list as arguments and return the position of the item in the list, like we saw before.
def indexlist(item2find, list_or_string):
"Returns all indexes of an item in a list or a string"
return [n for n,item in enumerate(list_or_string) if item==item2find]
print(indexlist("1", "010101010"))
Output
[1, 3, 5, 7]
Simple
for n, i in enumerate([1, 2, 3, 4, 1]):
if i == 1:
print(n)
Output:
0
4

One more solution(sorry if duplicates) for all occurrences:
values = [1,2,3,1,2,4,5,6,3,2,1]
map(lambda val: (val, [i for i in xrange(len(values)) if values[i] == val]), values)

Using filter() in python2.
>>> q = ['Yeehaw', 'Yeehaw', 'Googol', 'B9', 'Googol', 'NSM', 'B9', 'NSM', 'Dont Ask', 'Googol']
>>> filter(lambda i: q[i]=="Googol", range(len(q)))
[2, 4, 9]

If you need to search for all element's positions between certain indices, you can state them:
[i for i,x in enumerate([1,2,3,2]) if x==2 & 2<= i <=3] # -> [3]

A dynamic list comprehension based solution incase we do not know in advance which element:
lst = ['to', 'be', 'or', 'not', 'to', 'be']
{req_word: [idx for idx, word in enumerate(lst) if word == req_word] for req_word in set(lst)}
results in:
{'be': [1, 5], 'or': [2], 'to': [0, 4], 'not': [3]}
You can think of all other ways along the same lines as well but with index() you can find only one index although you can set occurrence number yourself.

Using a for-loop:
Answers with enumerate and a list comprehension are more pythonic, but not necessarily faster. However, this answer is aimed at students who may not be allowed to use some of those built-in functions.
create an empty list, indices
create the loop with for i in range(len(x)):, which essentially iterates through a list of index locations [0, 1, 2, 3, ..., len(x)-1]
in the loop, add any i, where x[i] is a match to value, to indices
x[i] accesses the list by index
def get_indices(x: list, value: int) -> list:
indices = list()
for i in range(len(x)):
if x[i] == value:
indices.append(i)
return indices
n = [1, 2, 3, -50, -60, 0, 6, 9, -60, -60]
print(get_indices(n, -60))
>>> [4, 8, 9]
The functions, get_indices, are implemented with type hints. In this case, the list, n, is a bunch of ints, therefore we search for value, also defined as an int.
Using a while-loop and .index:
With .index, use try-except for error handling, because a ValueError will occur if value is not in the list.
def get_indices(x: list, value: int) -> list:
indices = list()
i = 0
while True:
try:
# find an occurrence of value and update i to that index
i = x.index(value, i)
# add i to the list
indices.append(i)
# advance i by 1
i += 1
except ValueError as e:
break
return indices
print(get_indices(n, -60))
>>> [4, 8, 9]

You can create a defaultdict
from collections import defaultdict
d1 = defaultdict(int) # defaults to 0 values for keys
unq = set(lst1) # lst1 = [1, 2, 2, 3, 4, 1, 2, 7]
for each in unq:
d1[each] = lst1.count(each)
else:
print(d1)

If you are using Python 2, you can achieve the same functionality with this:
f = lambda my_list, value:filter(lambda x: my_list[x] == value, range(len(my_list)))
Where my_list is the list you want to get the indexes of, and value is the value searched. Usage:
f(some_list, some_element)

Create a generator
Generators are fast and use a tiny memory footprint. They give you flexibility in how you use the result.
def indices(iter, val):
"""Generator: Returns all indices of val in iter
Raises a ValueError if no val does not occur in iter
Passes on the AttributeError if iter does not have an index method (e.g. is a set)
"""
i = -1
NotFound = False
while not NotFound:
try:
i = iter.index(val, i+1)
except ValueError:
NotFound = True
else:
yield i
if i == -1:
raise ValueError("No occurrences of {v} in {i}".format(v = val, i = iter))
The above code can be use to create a list of the indices: list(indices(input,value)); use them as dictionary keys: dict(indices(input,value)); sum them: sum(indices(input,value)); in a for loop for index_ in indices(input,value):; ...etc... without creating an interim list/tuple or similar.
In a for loop you will get your next index back when you call for it, without waiting for all the others to be calculated first. That means: if you break out of the loop for some reason you save the time needed to find indices you never needed.
How it works
Call .index on the input iter to find the next occurrence of
val
Use the second parameter to .index to start at the point
after the last found occurrence
Yield the index
Repeat until index raises a ValueError
Alternative versions
I tried four different versions for flow control; two EAFP (using try - except) and two TBYL (with a logical test in the while statement):
"WhileTrueBreak": while True: ... except ValueError: break. Surprisingly, this was usually a touch slower than option 2 and (IMV) less readable
"WhileErrFalse": Using a bool variable err to identify when a ValueError is raised. This is generally the fastest and more readable than 1
"RemainingSlice": Check whether val is in the remaining part of the input using slicing: while val in iter[i:]. Unsurprisingly, this does not scale well
"LastOccurrence": Check first where the last occurrence is, keep going while i < last
The overall performance differences between 1,2 and 4 are negligible, so it comes down to personal style and preference. Given that .index uses ValueError to let you know it didn't find anything, rather than e.g. returning None, an EAFP-approach seems fitting to me.
Here are the 4 code variants and results from timeit (in milliseconds) for different lengths of input and sparsity of matches
#version("WhileTrueBreak", versions)
def indices2(iter, val):
i = -1
while True:
try:
i = iter.index(val, i+1)
except ValueError:
break
else:
yield i
#version("WhileErrFalse", versions)
def indices5(iter, val):
i = -1
err = False
while not err:
try:
i = iter.index(val, i+1)
except ValueError:
err = True
else:
yield i
#version("RemainingSlice", versions)
def indices1(iter, val):
i = 0
while val in iter[i:]:
i = iter.index(val, i)
yield i
i += 1
#version("LastOccurrence", versions)
def indices4(iter,val):
i = 0
last = len(iter) - tuple(reversed(iter)).index(val)
while i < last:
i = iter.index(val, i)
yield i
i += 1
Length: 100, Ocurrences: 4.0%
{'WhileTrueBreak': 0.0074799987487494946, 'WhileErrFalse': 0.006440002471208572, 'RemainingSlice': 0.01221001148223877, 'LastOccurrence': 0.00801000278443098}
Length: 1000, Ocurrences: 1.2%
{'WhileTrueBreak': 0.03101000329479575, 'WhileErrFalse': 0.0278000021353364, 'RemainingSlice': 0.08278000168502331, 'LastOccurrence': 0.03986000083386898}
Length: 10000, Ocurrences: 2.05%
{'WhileTrueBreak': 0.18062000162899494, 'WhileErrFalse': 0.1810499932616949, 'RemainingSlice': 2.9145700042136014, 'LastOccurrence': 0.2049500006251037}
Length: 100000, Ocurrences: 1.977%
{'WhileTrueBreak': 1.9361200043931603, 'WhileErrFalse': 1.7280600033700466, 'RemainingSlice': 254.4725100044161, 'LastOccurrence': 1.9101499929092824}
Length: 100000, Ocurrences: 9.873%
{'WhileTrueBreak': 2.832529996521771, 'WhileErrFalse': 2.9984100023284554, 'RemainingSlice': 1132.4922299943864, 'LastOccurrence': 2.6660699979402125}
Length: 100000, Ocurrences: 25.058%
{'WhileTrueBreak': 5.119729996658862, 'WhileErrFalse': 5.2082200068980455, 'RemainingSlice': 2443.0577100021765, 'LastOccurrence': 4.75954000139609}
Length: 100000, Ocurrences: 49.698%
{'WhileTrueBreak': 9.372120001353323, 'WhileErrFalse': 8.447749994229525, 'RemainingSlice': 5042.717969999649, 'LastOccurrence': 8.050809998530895}

Here is a time performance comparison between using np.where vs list_comprehension. Seems like np.where is faster on average.
# np.where
start_times = []
end_times = []
for i in range(10000):
start = time.time()
start_times.append(start)
temp_list = np.array([1,2,3,3,5])
ixs = np.where(temp_list==3)[0].tolist()
end = time.time()
end_times.append(end)
print("Took on average {} seconds".format(
np.mean(end_times)-np.mean(start_times)))
Took on average 3.81469726562e-06 seconds
# list_comprehension
start_times = []
end_times = []
for i in range(10000):
start = time.time()
start_times.append(start)
temp_list = np.array([1,2,3,3,5])
ixs = [i for i in range(len(temp_list)) if temp_list[i]==3]
end = time.time()
end_times.append(end)
print("Took on average {} seconds".format(
np.mean(end_times)-np.mean(start_times)))
Took on average 4.05311584473e-06 seconds

Python: Creating a function that compares 2 arrays and inserts the larger elements between the 2 arrays

I am trying to create a function that compares 2 arrays and creates a new list with the maximum elements of the list without using numpy. I managed to create a manual version, but am having issues implementing this into a function.
Task: Create a function maximum_arrays(a,b) that compares both arrays a and b element-wise and returns a new array containing the larger elements. Use the insert2 function to add new elements to a list.
Example: from applying the function to the arrays a=[12,5,8,19,6] and b=[3,6,2,12,4] the result should be c=[12,6,8,19,6].
Current Code:
list_a = [12,5,8,19,6]
list_b = [3,6,2,12,4]
maximum_arrays = []
for item in list_a:
if list_b[item] > list_a[item]:
maximum_arrays.insert(list_b[item])
else:
maximum_arrays.insert(list_a[item])
print(maximum_arrays)
Manual Version:
list_a = [12,5,8,19,6]
list_b = [3,6,2,12,4]
#answer example
c = [12,6,8,19,6]
#empty list
maximum_arrays = []
#for each part of the list, choose the highest number of the other list and insert
maximum_arrays.insert(0, max(list_a[0],list_b[0]))
maximum_arrays.insert(1, max(list_a[1],list_b[1]))
maximum_arrays.insert(2, max(list_a[2],list_b[2]))
maximum_arrays.insert(3, max(list_a[3],list_b[3]))
maximum_arrays.insert(4, max(list_a[4],list_b[4]))
print(maximum_arrays)

use max in a list comprehension over the zipped lists, or numpy.max.
list_a = [12,5,8,19,6]
list_b = [3,6,2,12,4]
max_array = [max(i) for i in zip(list_a, list_b)]
print(max_array)
The explanation here is: zip turns n iterables into an iterator over tuples, where each tuple has n items. So, in the two-list case, zip([1, 2, 3], [4, 5, 6]) turns into ((1, 4), (2, 5), (3, 6)). Taking the max of all of these tuples gives you your list.
An important caveat, and one that has burned me several times, is that the number of tuples generated is the length of the shortest iterable in the zip. In other words, zip does not throw an exception when passed iterables of different lengths, and just stops when one of the input lists runs out. In this respect it differs from numpy.max, which does throw an error when given lists of different lengths.

Are you looking for something like this :
list_a = [12,5,8,19,6]
list_b = [3,6,2,12,4]
l = []
for i,j in enumerate(zip(list_a, list_b)):
l.insert(i, max(j))
print(l)

Other way using itertools.starmap:
list(starmap(max, zip(list_a, list_b)))
Output:
[12, 6, 8, 19, 6]
This is about 1.4x faster than list comprehension:
%timeit list(starmap(max, zip(list_a, list_b)))
# 1.19 µs ± 49.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%timeit [max(i) for i in zip(list_a, list_b)]
# 1.69 µs ± 213 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

You can solve many ways , Same as your function or list comprehension using zip function.
Below solution on your way, If both length is same then take one list length and iterate and use the "append"/"insert" option to add the value to list.
list_a = [12,5,8,19,6]
list_b = [3,6,2,12,4]
maximum_arrays = []
list_length = len(list_a)
for item in range(list_length):
if list_b[item] > list_a[item]:
maximum_arrays.append(list_b[item])
else:
maximum_arrays.append(list_a[item])
print(maximum_arrays)

Find place of all same objects in list [duplicate]

index() will give the first occurrence of an item in a list. Is there a neat trick which returns all indices in a list for an element?

You can use a list comprehension with enumerate:
indices = [i for i, x in enumerate(my_list) if x == "whatever"]
The iterator enumerate(my_list) yields pairs (index, item) for each item in the list. Using i, x as loop variable target unpacks these pairs into the index i and the list item x. We filter down to all x that match our criterion, and select the indices i of these elements.

While not a solution for lists directly, numpy really shines for this sort of thing:
import numpy as np
values = np.array([1,2,3,1,2,4,5,6,3,2,1])
searchval = 3
ii = np.where(values == searchval)[0]
returns:
ii ==>array([2, 8])
This can be significantly faster for lists (arrays) with a large number of elements vs some of the other solutions.

A solution using list.index:
def indices(lst, element):
result = []
offset = -1
while True:
try:
offset = lst.index(element, offset+1)
except ValueError:
return result
result.append(offset)
It's much faster than the list comprehension with enumerate, for large lists. It is also much slower than the numpy solution if you already have the array, otherwise the cost of converting outweighs the speed gain (tested on integer lists with 100, 1000 and 10000 elements).
NOTE: A note of caution based on Chris_Rands' comment: this solution is faster than the list comprehension if the results are sufficiently sparse, but if the list has many instances of the element that is being searched (more than ~15% of the list, on a test with a list of 1000 integers), the list comprehension is faster.

How about:
In [1]: l=[1,2,3,4,3,2,5,6,7]
In [2]: [i for i,val in enumerate(l) if val==3]
Out[2]: [2, 4]

more_itertools.locate finds indices for all items that satisfy a condition.
from more_itertools import locate
list(locate([0, 1, 1, 0, 1, 0, 0]))
# [1, 2, 4]
list(locate(['a', 'b', 'c', 'b'], lambda x: x == 'b'))
# [1, 3]
more_itertools is a third-party library > pip install more_itertools.

occurrences = lambda s, lst: (i for i,e in enumerate(lst) if e == s)
list(occurrences(1, [1,2,3,1])) # = [0, 3]

Or Use range (python 3):
l=[i for i in range(len(lst)) if lst[i]=='something...']
For (python 2):
l=[i for i in xrange(len(lst)) if lst[i]=='something...']
And then (both cases):
print(l)
Is as expected.

There’s an answer using np.where to find the indices of a single value, which is not faster than a list-comprehension, if the time to convert a list to an array is included
The overhead of importing numpy and converting a list to a numpy.array probably makes using numpy a less efficient option for most circumstances. A careful timing analysis would be necessary.
In cases where multiple functions/operations will need to be performed on the list, converting the list to an array, and then using numpy functions will likely be a faster option.
This solution uses np.where and np.unique to find the indices of all unique elements in a list.
Using np.where on an array (including the time to convert the list to an array) is slightly slower than a list-comprehension on a list, for finding all indices of all unique elements.
This has been tested on an 2M element list with 4 unique values, and the size of the list/array and number of unique elements will have an impact.
Other solutions using numpy on an array can be found in Get a list of all indices of repeated elements in a numpy array
Tested in [python 3.10.4, numpy 1.23.1] and [python 3.11.0, numpy 1.23.4]
import numpy as np
import random # to create test list
# create sample list
random.seed(365)
l = [random.choice(['s1', 's2', 's3', 's4']) for _ in range(20)]
# convert the list to an array for use with these numpy methods
a = np.array(l)
# create a dict of each unique entry and the associated indices
idx = {v: np.where(a == v)[0].tolist() for v in np.unique(a)}
# print(idx)
{'s1': [7, 9, 10, 11, 17],
's2': [1, 3, 6, 8, 14, 18, 19],
's3': [0, 2, 13, 16],
's4': [4, 5, 12, 15]}
%timeit on a 2M element list with 4 unique str elements
# create 2M element list
random.seed(365)
l = [random.choice(['s1', 's2', 's3', 's4']) for _ in range(2000000)]
Functions
def test1():
# np.where: convert list to array and find indices of a single element
a = np.array(l)
return np.where(a == 's1')
def test2():
# list-comprehension: on list l and find indices of a single element
return [i for i, x in enumerate(l) if x == "s1"]
def test3():
# filter: on list l and find indices of a single element
return list(filter(lambda i: l[i]=="s1", range(len(l))))
def test4():
# use np.where and np.unique to find indices of all unique elements: convert list to array
a = np.array(l)
return {v: np.where(a == v)[0].tolist() for v in np.unique(a)}
def test5():
# list comprehension inside dict comprehension: on list l and find indices of all unique elements
return {req_word: [idx for idx, word in enumerate(l) if word == req_word] for req_word in set(l)}
Function Call
%timeit test1()
%timeit test2()
%timeit test3()
%timeit test4()
%timeit test5()
Result python 3.10.4
214 ms ± 19.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
85.1 ms ± 1.48 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
146 ms ± 1.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
365 ms ± 11.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
360 ms ± 5.82 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Result python 3.11.0
209 ms ± 15.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
70.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
132 ms ± 4.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
371 ms ± 20.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
314 ms ± 15.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Getting all the occurrences and the position of one or more (identical) items in a list
With enumerate(alist) you can store the first element (n) that is the index of the list when the element x is equal to what you look for.
>>> alist = ['foo', 'spam', 'egg', 'foo']
>>> foo_indexes = [n for n,x in enumerate(alist) if x=='foo']
>>> foo_indexes
[0, 3]
>>>
Let's make our function findindex
This function takes the item and the list as arguments and return the position of the item in the list, like we saw before.
def indexlist(item2find, list_or_string):
"Returns all indexes of an item in a list or a string"
return [n for n,item in enumerate(list_or_string) if item==item2find]
print(indexlist("1", "010101010"))
Output
[1, 3, 5, 7]
Simple
for n, i in enumerate([1, 2, 3, 4, 1]):
if i == 1:
print(n)
Output:
0
4

One more solution(sorry if duplicates) for all occurrences:
values = [1,2,3,1,2,4,5,6,3,2,1]
map(lambda val: (val, [i for i in xrange(len(values)) if values[i] == val]), values)

Using filter() in python2.
>>> q = ['Yeehaw', 'Yeehaw', 'Googol', 'B9', 'Googol', 'NSM', 'B9', 'NSM', 'Dont Ask', 'Googol']
>>> filter(lambda i: q[i]=="Googol", range(len(q)))
[2, 4, 9]

If you need to search for all element's positions between certain indices, you can state them:
[i for i,x in enumerate([1,2,3,2]) if x==2 & 2<= i <=3] # -> [3]

A dynamic list comprehension based solution incase we do not know in advance which element:
lst = ['to', 'be', 'or', 'not', 'to', 'be']
{req_word: [idx for idx, word in enumerate(lst) if word == req_word] for req_word in set(lst)}
results in:
{'be': [1, 5], 'or': [2], 'to': [0, 4], 'not': [3]}
You can think of all other ways along the same lines as well but with index() you can find only one index although you can set occurrence number yourself.

Using a for-loop:
Answers with enumerate and a list comprehension are more pythonic, but not necessarily faster. However, this answer is aimed at students who may not be allowed to use some of those built-in functions.
create an empty list, indices
create the loop with for i in range(len(x)):, which essentially iterates through a list of index locations [0, 1, 2, 3, ..., len(x)-1]
in the loop, add any i, where x[i] is a match to value, to indices
x[i] accesses the list by index
def get_indices(x: list, value: int) -> list:
indices = list()
for i in range(len(x)):
if x[i] == value:
indices.append(i)
return indices
n = [1, 2, 3, -50, -60, 0, 6, 9, -60, -60]
print(get_indices(n, -60))
>>> [4, 8, 9]
The functions, get_indices, are implemented with type hints. In this case, the list, n, is a bunch of ints, therefore we search for value, also defined as an int.
Using a while-loop and .index:
With .index, use try-except for error handling, because a ValueError will occur if value is not in the list.
def get_indices(x: list, value: int) -> list:
indices = list()
i = 0
while True:
try:
# find an occurrence of value and update i to that index
i = x.index(value, i)
# add i to the list
indices.append(i)
# advance i by 1
i += 1
except ValueError as e:
break
return indices
print(get_indices(n, -60))
>>> [4, 8, 9]

You can create a defaultdict
from collections import defaultdict
d1 = defaultdict(int) # defaults to 0 values for keys
unq = set(lst1) # lst1 = [1, 2, 2, 3, 4, 1, 2, 7]
for each in unq:
d1[each] = lst1.count(each)
else:
print(d1)

If you are using Python 2, you can achieve the same functionality with this:
f = lambda my_list, value:filter(lambda x: my_list[x] == value, range(len(my_list)))
Where my_list is the list you want to get the indexes of, and value is the value searched. Usage:
f(some_list, some_element)

Create a generator
Generators are fast and use a tiny memory footprint. They give you flexibility in how you use the result.
def indices(iter, val):
"""Generator: Returns all indices of val in iter
Raises a ValueError if no val does not occur in iter
Passes on the AttributeError if iter does not have an index method (e.g. is a set)
"""
i = -1
NotFound = False
while not NotFound:
try:
i = iter.index(val, i+1)
except ValueError:
NotFound = True
else:
yield i
if i == -1:
raise ValueError("No occurrences of {v} in {i}".format(v = val, i = iter))
The above code can be use to create a list of the indices: list(indices(input,value)); use them as dictionary keys: dict(indices(input,value)); sum them: sum(indices(input,value)); in a for loop for index_ in indices(input,value):; ...etc... without creating an interim list/tuple or similar.
In a for loop you will get your next index back when you call for it, without waiting for all the others to be calculated first. That means: if you break out of the loop for some reason you save the time needed to find indices you never needed.
How it works
Call .index on the input iter to find the next occurrence of
val
Use the second parameter to .index to start at the point
after the last found occurrence
Yield the index
Repeat until index raises a ValueError
Alternative versions
I tried four different versions for flow control; two EAFP (using try - except) and two TBYL (with a logical test in the while statement):
"WhileTrueBreak": while True: ... except ValueError: break. Surprisingly, this was usually a touch slower than option 2 and (IMV) less readable
"WhileErrFalse": Using a bool variable err to identify when a ValueError is raised. This is generally the fastest and more readable than 1
"RemainingSlice": Check whether val is in the remaining part of the input using slicing: while val in iter[i:]. Unsurprisingly, this does not scale well
"LastOccurrence": Check first where the last occurrence is, keep going while i < last
The overall performance differences between 1,2 and 4 are negligible, so it comes down to personal style and preference. Given that .index uses ValueError to let you know it didn't find anything, rather than e.g. returning None, an EAFP-approach seems fitting to me.
Here are the 4 code variants and results from timeit (in milliseconds) for different lengths of input and sparsity of matches
#version("WhileTrueBreak", versions)
def indices2(iter, val):
i = -1
while True:
try:
i = iter.index(val, i+1)
except ValueError:
break
else:
yield i
#version("WhileErrFalse", versions)
def indices5(iter, val):
i = -1
err = False
while not err:
try:
i = iter.index(val, i+1)
except ValueError:
err = True
else:
yield i
#version("RemainingSlice", versions)
def indices1(iter, val):
i = 0
while val in iter[i:]:
i = iter.index(val, i)
yield i
i += 1
#version("LastOccurrence", versions)
def indices4(iter,val):
i = 0
last = len(iter) - tuple(reversed(iter)).index(val)
while i < last:
i = iter.index(val, i)
yield i
i += 1
Length: 100, Ocurrences: 4.0%
{'WhileTrueBreak': 0.0074799987487494946, 'WhileErrFalse': 0.006440002471208572, 'RemainingSlice': 0.01221001148223877, 'LastOccurrence': 0.00801000278443098}
Length: 1000, Ocurrences: 1.2%
{'WhileTrueBreak': 0.03101000329479575, 'WhileErrFalse': 0.0278000021353364, 'RemainingSlice': 0.08278000168502331, 'LastOccurrence': 0.03986000083386898}
Length: 10000, Ocurrences: 2.05%
{'WhileTrueBreak': 0.18062000162899494, 'WhileErrFalse': 0.1810499932616949, 'RemainingSlice': 2.9145700042136014, 'LastOccurrence': 0.2049500006251037}
Length: 100000, Ocurrences: 1.977%
{'WhileTrueBreak': 1.9361200043931603, 'WhileErrFalse': 1.7280600033700466, 'RemainingSlice': 254.4725100044161, 'LastOccurrence': 1.9101499929092824}
Length: 100000, Ocurrences: 9.873%
{'WhileTrueBreak': 2.832529996521771, 'WhileErrFalse': 2.9984100023284554, 'RemainingSlice': 1132.4922299943864, 'LastOccurrence': 2.6660699979402125}
Length: 100000, Ocurrences: 25.058%
{'WhileTrueBreak': 5.119729996658862, 'WhileErrFalse': 5.2082200068980455, 'RemainingSlice': 2443.0577100021765, 'LastOccurrence': 4.75954000139609}
Length: 100000, Ocurrences: 49.698%
{'WhileTrueBreak': 9.372120001353323, 'WhileErrFalse': 8.447749994229525, 'RemainingSlice': 5042.717969999649, 'LastOccurrence': 8.050809998530895}

Here is a time performance comparison between using np.where vs list_comprehension. Seems like np.where is faster on average.
# np.where
start_times = []
end_times = []
for i in range(10000):
start = time.time()
start_times.append(start)
temp_list = np.array([1,2,3,3,5])
ixs = np.where(temp_list==3)[0].tolist()
end = time.time()
end_times.append(end)
print("Took on average {} seconds".format(
np.mean(end_times)-np.mean(start_times)))
Took on average 3.81469726562e-06 seconds
# list_comprehension
start_times = []
end_times = []
for i in range(10000):
start = time.time()
start_times.append(start)
temp_list = np.array([1,2,3,3,5])
ixs = [i for i in range(len(temp_list)) if temp_list[i]==3]
end = time.time()
end_times.append(end)
print("Took on average {} seconds".format(
np.mean(end_times)-np.mean(start_times)))
Took on average 4.05311584473e-06 seconds

How do a split integers within a list into single digits only?

Let's say I have something like this:
list(range(9:12))
Which gives me a list:
[9,10,11]
However I want it to be like:
[9,1,0,1,1]
Which splits every integer into single digits, is there anyway of achieving this without sacrificing too much performance? Or is there a way of generating list like these in the first place?

You can build the final result efficiently without having to build one large and/or small intermediate strings using itertools.chain.from_iterable.
In [18]: list(map(int, chain.from_iterable(map(str, range(9, 12)))))
Out[18]: [9, 1, 0, 1, 1]
In [12]: %%timeit
...: list(map(int, chain.from_iterable(map(str, range(9, 20)))))
...:
100000 loops, best of 3: 8.19 µs per loop
In [13]: %%timeit
...: [int(i) for i in ''.join(map(str, range(9, 20)))]
...:
100000 loops, best of 3: 9.15 µs per loop
In [14]: %%timeit
...: [int(x) for i in range(9, 20) for x in str(i)]
...:
100000 loops, best of 3: 9.92 µs per loop
Timings scale with input. The itertools version also uses memory efficiently although it is marginally slower than the str.join version if used with list(map(int, ...)):
In [15]: %%timeit
...: list(map(int, chain.from_iterable(map(str, range(9, 200)))))
...:
10000 loops, best of 3: 138 µs per loop
In [16]: %%timeit
...: [int(i) for i in ''.join(map(str, range(9, 200)))]
...:
10000 loops, best of 3: 159 µs per loop
In [17]: %%timeit
...: [int(x) for i in range(9, 200) for x in str(i)]
...:
10000 loops, best of 3: 182 µs per loop
In [18]: %%timeit
...: list(map(int, ''.join(map(str, range(9, 200)))))
...:
10000 loops, best of 3: 130 µs per loop

simplest way is,
>>> [int(i) for i in range(9,12) for i in str(i)]
[9, 1, 0, 1, 1]
>>>

Convert the integers to strings, then split() the string and reconvert the digits back to ints.
li = range(9,12)
digitlist = [int(d) for number in li for d in str(number)]
Output:
[9,1,0,1,1]

I've investigated how performant I can make this a little more. The first function I wrote was naive_single_digits, which uses the str approach, with a pretty efficient list comprehension.
def naive_single_digits(l):
return [int(c) for n in l
for c in str(n)]
As you can see, this approach works:
In [2]: naive_single_digits(range(9, 15))
Out[2]: [9, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4]
However, I thought that it would surely be unecessary to always build a str object for each item in the list - all we actually need is a base conversion to digits. Out of laziness, I copied this function from here. I've optimised it a bit by specifying it to base 10.
def base10(n):
if n == 0:
return [0]
digits = []
while n:
digits.append(n % 10)
n //= 10
return digits[::-1]
Using this, I made
def arithmetic_single_digits(l):
return [i for n in l
for i in base10(n)]
which also behaves correctly:
In [3]: arithmetic_single_digits(range(9, 15))
Out[3]: [9, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4]
Now to time it. I've also tested against one other answer (full disclosure: I modified it a bit to work in Python2, but that shouldn't have affected the performance much)
In [11]: %timeit -n 10 naive_single_digits(range(100000))
10 loops, best of 3: 173 ms per loop
In [10]: %timeit -n 10 list(map(int, itertools.chain(*map(str, range(100000)))))
10 loops, best of 3: 154 ms per loop
In [12]: %timeit arithmetic_single_digits(range(100000))
10 loops, best of 3: 93.3 ms per loop
As you can see arithmetic_single_digits is actually somewhat faster, although this is at the cost of more code and possibly less clarity. I've tested against ridiculously large inputs, so you can see a difference in performance - at any kind of reasonable scale, every answer here will be blazingly fast. Note that python's integer arithmetic is probably actually relatively slow, as it doesn't use a primitive integer type. If this were to be implemented in C, I'd suspect my approach to get a bit faster.
Comparing this to viblo's answer, using (pure) Python 3 (to my shame I haven't installed ipython for python 3):
print(timeit.timeit("digits(range(1, 100000))", number=10, globals=globals()))
print(timeit.timeit("arithmetic_single_digits(range(1, 100000))", number=10, globals=globals()))
This has the output of:
3.5284318959747907
0.806847038998967
My approach is quite a bit faster, presumably because I'm purely using integer arithmetic.

Another way to write an arithmetic solution. Compared to Izaak van Dongens solution this doesnt use a while loop but calculates upfront how many iterations it need in the list comprehension/loop.
import itertools, math
def digits(ns):
return list(itertools.chain.from_iterable(
[
[
(abs(n) - (abs(n) // 10 **x)*10**x ) // 10**(x-1)
for x
in range(1+math.floor(math.log10(abs(n) if n!=0 else 1)), 0, -1)]
for n in ns
]
))
digits([-11,-10,-9,0,9,10,11])

Turn it to a string then back into a list :)
lambda x: list(''.join(str(e) for e in x))

You can also do with map function
a=range(9,12)
res = []
b=[map(int, str(i)) for i in a]
for i in b:
res.extend(i)
print(res)

here is how I did it:
ls = range(9,12)
lsNew = []
length = len(ls)
for i in range(length):
item = ls[i]
string = str(item)
if len(string) > 1:
split = list(string)
lsNew = lsNew + split
else:
lsNew.append(item)
ls = lsNew
print(ls)

def breakall(L):
if L == []:
return []
elif L[0] < 10:
return [L[0]] + breakall(L[1:])
else:
return breakall([L[0]//10]) + [L[0] % 10] + breakall(L[1:])
print(breakall([9,10,12]))
-->
[9, 1, 0, 1, 2]