Related
Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index 1?
>>> ["foo", "bar", "baz"].index("bar")
1
See the documentation for the built-in .index() method of the list:
list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.
The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.
Caveats
Linear time-complexity in list length
An index call checks every element of the list in order, until it finds a match. If the list is long, and if there is no guarantee that the value will be near the beginning, this can slow down the code.
This problem can only be completely avoided by using a different data structure. However, if the element is known to be within a certain part of the list, the start and end parameters can be used to narrow the search.
For example:
>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514
The second call is orders of magnitude faster, because it only has to search through 10 elements, rather than all 1 million.
Only the index of the first match is returned
A call to index searches through the list in order until it finds a match, and stops there. If there could be more than one occurrence of the value, and all indices are needed, index cannot solve the problem:
>>> [1, 1].index(1) # the `1` index is not found.
0
Instead, use a list comprehension or generator expression to do the search, with enumerate to get indices:
>>> # A list comprehension gives a list of indices directly:
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> # A generator comprehension gives us an iterable object...
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> # which can be used in a `for` loop, or manually iterated with `next`:
>>> next(g)
0
>>> next(g)
2
The list comprehension and generator expression techniques still work if there is only one match, and are more generalizable.
Raises an exception if there is no match
As noted in the documentation above, using .index will raise an exception if the searched-for value is not in the list:
>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
If this is a concern, either explicitly check first using item in my_list, or handle the exception with try/except as appropriate.
The explicit check is simple and readable, but it must iterate the list a second time. See What is the EAFP principle in Python? for more guidance on this choice.
The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():
for i, j in enumerate(['foo', 'bar', 'baz']):
if j == 'bar':
print(i)
The index() function only returns the first occurrence, while enumerate() returns all occurrences.
As a list comprehension:
[i for i, j in enumerate(['foo', 'bar', 'baz']) if j == 'bar']
Here's also another small solution with itertools.count() (which is pretty much the same approach as enumerate):
from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ['foo', 'bar', 'baz']) if j == 'bar']
This is more efficient for larger lists than using enumerate():
$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ['foo', 'bar', 'baz']*500) if j == 'bar']"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(['foo', 'bar', 'baz']*500) if j == 'bar']"
10000 loops, best of 3: 196 usec per loop
To get all indexes:
indexes = [i for i, x in enumerate(xs) if x == 'foo']
index() returns the first index of value!
| index(...)
| L.index(value, [start, [stop]]) -> integer -- return first index of value
def all_indices(value, qlist):
indices = []
idx = -1
while True:
try:
idx = qlist.index(value, idx+1)
indices.append(idx)
except ValueError:
break
return indices
all_indices("foo", ["foo","bar","baz","foo"])
A problem will arise if the element is not in the list. This function handles the issue:
# if element is found it returns index of element else returns None
def find_element_in_list(element, list_element):
try:
index_element = list_element.index(element)
return index_element
except ValueError:
return None
a = ["foo","bar","baz",'bar','any','much']
indexes = [index for index in range(len(a)) if a[index] == 'bar']
You have to set a condition to check if the element you're searching is in the list
if 'your_element' in mylist:
print mylist.index('your_element')
else:
print None
If you want all indexes, then you can use NumPy:
import numpy as np
array = [1, 2, 1, 3, 4, 5, 1]
item = 1
np_array = np.array(array)
item_index = np.where(np_array==item)
print item_index
# Out: (array([0, 2, 6], dtype=int64),)
It is clear, readable solution.
All of the proposed functions here reproduce inherent language behavior but obscure what's going on.
[i for i in range(len(mylist)) if mylist[i]==myterm] # get the indices
[each for each in mylist if each==myterm] # get the items
mylist.index(myterm) if myterm in mylist else None # get the first index and fail quietly
Why write a function with exception handling if the language provides the methods to do what you want itself?
Finding the index of an item given a list containing it in Python
For a list ["foo", "bar", "baz"] and an item in the list "bar", what's the cleanest way to get its index (1) in Python?
Well, sure, there's the index method, which returns the index of the first occurrence:
>>> l = ["foo", "bar", "baz"]
>>> l.index('bar')
1
There are a couple of issues with this method:
if the value isn't in the list, you'll get a ValueError
if more than one of the value is in the list, you only get the index for the first one
No values
If the value could be missing, you need to catch the ValueError.
You can do so with a reusable definition like this:
def index(a_list, value):
try:
return a_list.index(value)
except ValueError:
return None
And use it like this:
>>> print(index(l, 'quux'))
None
>>> print(index(l, 'bar'))
1
And the downside of this is that you will probably have a check for if the returned value is or is not None:
result = index(a_list, value)
if result is not None:
do_something(result)
More than one value in the list
If you could have more occurrences, you'll not get complete information with list.index:
>>> l.append('bar')
>>> l
['foo', 'bar', 'baz', 'bar']
>>> l.index('bar') # nothing at index 3?
1
You might enumerate into a list comprehension the indexes:
>>> [index for index, v in enumerate(l) if v == 'bar']
[1, 3]
>>> [index for index, v in enumerate(l) if v == 'boink']
[]
If you have no occurrences, you can check for that with boolean check of the result, or just do nothing if you loop over the results:
indexes = [index for index, v in enumerate(l) if v == 'boink']
for index in indexes:
do_something(index)
Better data munging with pandas
If you have pandas, you can easily get this information with a Series object:
>>> import pandas as pd
>>> series = pd.Series(l)
>>> series
0 foo
1 bar
2 baz
3 bar
dtype: object
A comparison check will return a series of booleans:
>>> series == 'bar'
0 False
1 True
2 False
3 True
dtype: bool
Pass that series of booleans to the series via subscript notation, and you get just the matching members:
>>> series[series == 'bar']
1 bar
3 bar
dtype: object
If you want just the indexes, the index attribute returns a series of integers:
>>> series[series == 'bar'].index
Int64Index([1, 3], dtype='int64')
And if you want them in a list or tuple, just pass them to the constructor:
>>> list(series[series == 'bar'].index)
[1, 3]
Yes, you could use a list comprehension with enumerate too, but that's just not as elegant, in my opinion - you're doing tests for equality in Python, instead of letting builtin code written in C handle it:
>>> [i for i, value in enumerate(l) if value == 'bar']
[1, 3]
Is this an XY problem?
The XY problem is asking about your attempted solution rather than your actual problem.
Why do you think you need the index given an element in a list?
If you already know the value, why do you care where it is in a list?
If the value isn't there, catching the ValueError is rather verbose - and I prefer to avoid that.
I'm usually iterating over the list anyways, so I'll usually keep a pointer to any interesting information, getting the index with enumerate.
If you're munging data, you should probably be using pandas - which has far more elegant tools than the pure Python workarounds I've shown.
I do not recall needing list.index, myself. However, I have looked through the Python standard library, and I see some excellent uses for it.
There are many, many uses for it in idlelib, for GUI and text parsing.
The keyword module uses it to find comment markers in the module to automatically regenerate the list of keywords in it via metaprogramming.
In Lib/mailbox.py it seems to be using it like an ordered mapping:
key_list[key_list.index(old)] = new
and
del key_list[key_list.index(key)]
In Lib/http/cookiejar.py, seems to be used to get the next month:
mon = MONTHS_LOWER.index(mon.lower())+1
In Lib/tarfile.py similar to distutils to get a slice up to an item:
members = members[:members.index(tarinfo)]
In Lib/pickletools.py:
numtopop = before.index(markobject)
What these usages seem to have in common is that they seem to operate on lists of constrained sizes (important because of O(n) lookup time for list.index), and they're mostly used in parsing (and UI in the case of Idle).
While there are use-cases for it, they are fairly uncommon. If you find yourself looking for this answer, ask yourself if what you're doing is the most direct usage of the tools provided by the language for your use-case.
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
me = ["foo", "bar", "baz"]
me.index("bar")
You can apply this for any member of the list to get their index
All indexes with the zip function:
get_indexes = lambda x, xs: [i for (y, i) in zip(xs, range(len(xs))) if x == y]
print get_indexes(2, [1, 2, 3, 4, 5, 6, 3, 2, 3, 2])
print get_indexes('f', 'xsfhhttytffsafweef')
Simply you can go with
a = [['hand', 'head'], ['phone', 'wallet'], ['lost', 'stock']]
b = ['phone', 'lost']
res = [[x[0] for x in a].index(y) for y in b]
Another option
>>> a = ['red', 'blue', 'green', 'red']
>>> b = 'red'
>>> offset = 0;
>>> indices = list()
>>> for i in range(a.count(b)):
... indices.append(a.index(b,offset))
... offset = indices[-1]+1
...
>>> indices
[0, 3]
>>>
And now, for something completely different...
... like confirming the existence of the item before getting the index. The nice thing about this approach is the function always returns a list of indices -- even if it is an empty list. It works with strings as well.
def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
retval = []
last = 0
while val in l[last:]:
i = l[last:].index(val)
retval.append(last + i)
last += i + 1
return retval
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
When pasted into an interactive python window:
Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(the_list, val):
... """Always returns a list containing the indices of val in the_list"""
... retval = []
... last = 0
... while val in the_list[last:]:
... i = the_list[last:].index(val)
... retval.append(last + i)
... last += i + 1
... return retval
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
Update
After another year of heads-down python development, I'm a bit embarrassed by my original answer, so to set the record straight, one can certainly use the above code; however, the much more idiomatic way to get the same behavior would be to use list comprehension, along with the enumerate() function.
Something like this:
def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
return [index for index, value in enumerate(l) if value == val]
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
Which, when pasted into an interactive python window yields:
Python 2.7.14 |Anaconda, Inc.| (default, Dec 7 2017, 11:07:58)
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(l, val):
... """Always returns a list containing the indices of val in the_list"""
... return [index for index, value in enumerate(l) if value == val]
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
And now, after reviewing this question and all the answers, I realize that this is exactly what FMc suggested in his earlier answer. At the time I originally answered this question, I didn't even see that answer, because I didn't understand it. I hope that my somewhat more verbose example will aid understanding.
If the single line of code above still doesn't make sense to you, I highly recommend you Google 'python list comprehension' and take a few minutes to familiarize yourself. It's just one of the many powerful features that make it a joy to use Python to develop code.
Here's a two-liner using Python's index() function:
LIST = ['foo' ,'boo', 'shoo']
print(LIST.index('boo'))
Output: 1
A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry:
>>> a = ['foo','bar','baz','bar','any', 'foo', 'much']
>>> l = dict(zip(set(a), map(lambda y: [i for i,z in enumerate(a) if z is y ], set(a))))
>>> l['foo']
[0, 5]
>>> l ['much']
[6]
>>> l
{'baz': [2], 'foo': [0, 5], 'bar': [1, 3], 'any': [4], 'much': [6]}
>>>
You could also use this as a one liner to get all indices for a single entry. There are no guarantees for efficiency, though I did use set(a) to reduce the number of times the lambda is called.
Finding index of item x in list L:
idx = L.index(x) if (x in L) else -1
This solution is not as powerful as others, but if you're a beginner and only know about forloops it's still possible to find the first index of an item while avoiding the ValueError:
def find_element(p,t):
i = 0
for e in p:
if e == t:
return i
else:
i +=1
return -1
There is a chance that that value may not be present so to avoid this ValueError, we can check if that actually exists in the list .
list = ["foo", "bar", "baz"]
item_to_find = "foo"
if item_to_find in list:
index = list.index(item_to_find)
print("Index of the item is " + str(index))
else:
print("That word does not exist")
List comprehension would be the best option to acquire a compact implementation in finding the index of an item in a list.
a_list = ["a", "b", "a"]
print([index for (index , item) in enumerate(a_list) if item == "a"])
It just uses the python function array.index() and with a simple Try / Except it returns the position of the record if it is found in the list and return -1 if it is not found in the list (like on JavaScript with the function indexOf()).
fruits = ['apple', 'banana', 'cherry']
try:
pos = fruits.index("mango")
except:
pos = -1
In this case "mango" is not present in the list fruits so the pos variable is -1, if I had searched for "cherry" the pos variable would be 2.
There is a more functional answer to this.
list(filter(lambda x: x[1]=="bar",enumerate(["foo", "bar", "baz", "bar", "baz", "bar", "a", "b", "c"])))
More generic form:
def get_index_of(lst, element):
return list(map(lambda x: x[0],\
(list(filter(lambda x: x[1]==element, enumerate(lst))))))
For one comparable
# Throws ValueError if nothing is found
some_list = ['foo', 'bar', 'baz'].index('baz')
# some_list == 2
Custom predicate
some_list = [item1, item2, item3]
# Throws StopIteration if nothing is found
# *unless* you provide a second parameter to `next`
index_of_value_you_like = next(
i for i, item in enumerate(some_list)
if item.matches_your_criteria())
Finding index of all items by predicate
index_of_staff_members = [
i for i, user in enumerate(users)
if user.is_staff()]
Python index() method throws an error if the item was not found. So instead you can make it similar to the indexOf() function of JavaScript which returns -1 if the item was not found:
try:
index = array.index('search_keyword')
except ValueError:
index = -1
name ="bar"
list = [["foo", 1], ["bar", 2], ["baz", 3]]
new_list=[]
for item in list:
new_list.append(item[0])
print(new_list)
try:
location= new_list.index(name)
except:
location=-1
print (location)
This accounts for if the string is not in the list too, if it isn't in the list then location = -1
If you are going to find an index once then using "index" method is fine. However, if you are going to search your data more than once then I recommend using bisect module. Keep in mind that using bisect module data must be sorted. So you sort data once and then you can use bisect.
Using bisect module on my machine is about 20 times faster than using index method.
Here is an example of code using Python 3.8 and above syntax:
import bisect
from timeit import timeit
def bisect_search(container, value):
return (
index
if (index := bisect.bisect_left(container, value)) < len(container)
and container[index] == value else -1
)
data = list(range(1000))
# value to search
value = 666
# times to test
ttt = 1000
t1 = timeit(lambda: data.index(value), number=ttt)
t2 = timeit(lambda: bisect_search(data, value), number=ttt)
print(f"{t1=:.4f}, {t2=:.4f}, diffs {t1/t2=:.2f}")
Output:
t1=0.0400, t2=0.0020, diffs t1/t2=19.60
For those coming from another language like me, maybe with a simple loop it's easier to understand and use it:
mylist = ["foo", "bar", "baz", "bar"]
newlist = enumerate(mylist)
for index, item in newlist:
if item == "bar":
print(index, item)
I am thankful for So what exactly does enumerate do?. That helped me to understand.
Since Python lists are zero-based, we can use the zip built-in function as follows:
>>> [i for i,j in zip(range(len(haystack)), haystack) if j == 'needle' ]
where "haystack" is the list in question and "needle" is the item to look for.
(Note: Here we are iterating using i to get the indexes, but if we need rather to focus on the items we can switch to j.)
In PyCharm, when I write:
return set([(sy + ady, sx + adx)])
it says "Function call can be replaced with set literal" so it replaces it with:
return {(sy + ady, sx + adx)}
Why is that? A set() in Python is not the same as a dictionary {}?
And if it wants to optimize this, why is this more effective?
Python sets and dictionaries can both be constructed using curly braces:
my_dict = {'a': 1, 'b': 2}
my_set = {1, 2, 3}
The interpreter (and human readers) can distinguish between them based on their contents. However it isn't possible to distinguish between an empty set and an empty dict, so this case you need to use set() for empty sets to disambiguate.
A very simple test suggests that the literal construction is faster (python3.5):
>>> timeit.timeit('a = set([1, 2, 3])')
0.5449375328607857
>>> timeit.timeit('a = {1, 2, 3}')
0.20525191631168127
This question covers some issues of performance of literal constructions over builtin functions, albeit for lists and dicts. The summary seems to be that literal constructions require less work from the interpreter.
It is an alternative syntax for set()
>>> a = {1, 2}
>>> b = set()
>>> b.add(1)
>>> b.add(2)
>>> b
set([1, 2])
>>> a
set([1, 2])
>>> a == b
True
>>> type(a) == type(b)
True
dict syntax is different. It consists of key-value pairs. For example:
my_obj = {1:None, 2:None}
Another example how set and {} are not interchangeable (as jonrsharpe mentioned):
In: f = 'FH'
In: set(f)
Out: {'F', 'H'}
In: {f}
Out: {'FH'}
set([iterable]) is the constructor to create a set from the optional iterable iterable. And {} is to create set / dict object literals. So what is created depends on how you use it.
In [414]: x = {}
In [415]: type(x)
Out[415]: dict
In [416]: x = {1}
In [417]: type(x)
Out[417]: set
In [418]: x = {1: "hello"}
In [419]: type(x)
Out[419]: dict
Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index 1?
>>> ["foo", "bar", "baz"].index("bar")
1
See the documentation for the built-in .index() method of the list:
list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.
The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.
Caveats
Linear time-complexity in list length
An index call checks every element of the list in order, until it finds a match. If the list is long, and if there is no guarantee that the value will be near the beginning, this can slow down the code.
This problem can only be completely avoided by using a different data structure. However, if the element is known to be within a certain part of the list, the start and end parameters can be used to narrow the search.
For example:
>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514
The second call is orders of magnitude faster, because it only has to search through 10 elements, rather than all 1 million.
Only the index of the first match is returned
A call to index searches through the list in order until it finds a match, and stops there. If there could be more than one occurrence of the value, and all indices are needed, index cannot solve the problem:
>>> [1, 1].index(1) # the `1` index is not found.
0
Instead, use a list comprehension or generator expression to do the search, with enumerate to get indices:
>>> # A list comprehension gives a list of indices directly:
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> # A generator comprehension gives us an iterable object...
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> # which can be used in a `for` loop, or manually iterated with `next`:
>>> next(g)
0
>>> next(g)
2
The list comprehension and generator expression techniques still work if there is only one match, and are more generalizable.
Raises an exception if there is no match
As noted in the documentation above, using .index will raise an exception if the searched-for value is not in the list:
>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
If this is a concern, either explicitly check first using item in my_list, or handle the exception with try/except as appropriate.
The explicit check is simple and readable, but it must iterate the list a second time. See What is the EAFP principle in Python? for more guidance on this choice.
The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():
for i, j in enumerate(['foo', 'bar', 'baz']):
if j == 'bar':
print(i)
The index() function only returns the first occurrence, while enumerate() returns all occurrences.
As a list comprehension:
[i for i, j in enumerate(['foo', 'bar', 'baz']) if j == 'bar']
Here's also another small solution with itertools.count() (which is pretty much the same approach as enumerate):
from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ['foo', 'bar', 'baz']) if j == 'bar']
This is more efficient for larger lists than using enumerate():
$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ['foo', 'bar', 'baz']*500) if j == 'bar']"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(['foo', 'bar', 'baz']*500) if j == 'bar']"
10000 loops, best of 3: 196 usec per loop
To get all indexes:
indexes = [i for i, x in enumerate(xs) if x == 'foo']
index() returns the first index of value!
| index(...)
| L.index(value, [start, [stop]]) -> integer -- return first index of value
def all_indices(value, qlist):
indices = []
idx = -1
while True:
try:
idx = qlist.index(value, idx+1)
indices.append(idx)
except ValueError:
break
return indices
all_indices("foo", ["foo","bar","baz","foo"])
A problem will arise if the element is not in the list. This function handles the issue:
# if element is found it returns index of element else returns None
def find_element_in_list(element, list_element):
try:
index_element = list_element.index(element)
return index_element
except ValueError:
return None
a = ["foo","bar","baz",'bar','any','much']
indexes = [index for index in range(len(a)) if a[index] == 'bar']
You have to set a condition to check if the element you're searching is in the list
if 'your_element' in mylist:
print mylist.index('your_element')
else:
print None
If you want all indexes, then you can use NumPy:
import numpy as np
array = [1, 2, 1, 3, 4, 5, 1]
item = 1
np_array = np.array(array)
item_index = np.where(np_array==item)
print item_index
# Out: (array([0, 2, 6], dtype=int64),)
It is clear, readable solution.
All of the proposed functions here reproduce inherent language behavior but obscure what's going on.
[i for i in range(len(mylist)) if mylist[i]==myterm] # get the indices
[each for each in mylist if each==myterm] # get the items
mylist.index(myterm) if myterm in mylist else None # get the first index and fail quietly
Why write a function with exception handling if the language provides the methods to do what you want itself?
Finding the index of an item given a list containing it in Python
For a list ["foo", "bar", "baz"] and an item in the list "bar", what's the cleanest way to get its index (1) in Python?
Well, sure, there's the index method, which returns the index of the first occurrence:
>>> l = ["foo", "bar", "baz"]
>>> l.index('bar')
1
There are a couple of issues with this method:
if the value isn't in the list, you'll get a ValueError
if more than one of the value is in the list, you only get the index for the first one
No values
If the value could be missing, you need to catch the ValueError.
You can do so with a reusable definition like this:
def index(a_list, value):
try:
return a_list.index(value)
except ValueError:
return None
And use it like this:
>>> print(index(l, 'quux'))
None
>>> print(index(l, 'bar'))
1
And the downside of this is that you will probably have a check for if the returned value is or is not None:
result = index(a_list, value)
if result is not None:
do_something(result)
More than one value in the list
If you could have more occurrences, you'll not get complete information with list.index:
>>> l.append('bar')
>>> l
['foo', 'bar', 'baz', 'bar']
>>> l.index('bar') # nothing at index 3?
1
You might enumerate into a list comprehension the indexes:
>>> [index for index, v in enumerate(l) if v == 'bar']
[1, 3]
>>> [index for index, v in enumerate(l) if v == 'boink']
[]
If you have no occurrences, you can check for that with boolean check of the result, or just do nothing if you loop over the results:
indexes = [index for index, v in enumerate(l) if v == 'boink']
for index in indexes:
do_something(index)
Better data munging with pandas
If you have pandas, you can easily get this information with a Series object:
>>> import pandas as pd
>>> series = pd.Series(l)
>>> series
0 foo
1 bar
2 baz
3 bar
dtype: object
A comparison check will return a series of booleans:
>>> series == 'bar'
0 False
1 True
2 False
3 True
dtype: bool
Pass that series of booleans to the series via subscript notation, and you get just the matching members:
>>> series[series == 'bar']
1 bar
3 bar
dtype: object
If you want just the indexes, the index attribute returns a series of integers:
>>> series[series == 'bar'].index
Int64Index([1, 3], dtype='int64')
And if you want them in a list or tuple, just pass them to the constructor:
>>> list(series[series == 'bar'].index)
[1, 3]
Yes, you could use a list comprehension with enumerate too, but that's just not as elegant, in my opinion - you're doing tests for equality in Python, instead of letting builtin code written in C handle it:
>>> [i for i, value in enumerate(l) if value == 'bar']
[1, 3]
Is this an XY problem?
The XY problem is asking about your attempted solution rather than your actual problem.
Why do you think you need the index given an element in a list?
If you already know the value, why do you care where it is in a list?
If the value isn't there, catching the ValueError is rather verbose - and I prefer to avoid that.
I'm usually iterating over the list anyways, so I'll usually keep a pointer to any interesting information, getting the index with enumerate.
If you're munging data, you should probably be using pandas - which has far more elegant tools than the pure Python workarounds I've shown.
I do not recall needing list.index, myself. However, I have looked through the Python standard library, and I see some excellent uses for it.
There are many, many uses for it in idlelib, for GUI and text parsing.
The keyword module uses it to find comment markers in the module to automatically regenerate the list of keywords in it via metaprogramming.
In Lib/mailbox.py it seems to be using it like an ordered mapping:
key_list[key_list.index(old)] = new
and
del key_list[key_list.index(key)]
In Lib/http/cookiejar.py, seems to be used to get the next month:
mon = MONTHS_LOWER.index(mon.lower())+1
In Lib/tarfile.py similar to distutils to get a slice up to an item:
members = members[:members.index(tarinfo)]
In Lib/pickletools.py:
numtopop = before.index(markobject)
What these usages seem to have in common is that they seem to operate on lists of constrained sizes (important because of O(n) lookup time for list.index), and they're mostly used in parsing (and UI in the case of Idle).
While there are use-cases for it, they are fairly uncommon. If you find yourself looking for this answer, ask yourself if what you're doing is the most direct usage of the tools provided by the language for your use-case.
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
me = ["foo", "bar", "baz"]
me.index("bar")
You can apply this for any member of the list to get their index
All indexes with the zip function:
get_indexes = lambda x, xs: [i for (y, i) in zip(xs, range(len(xs))) if x == y]
print get_indexes(2, [1, 2, 3, 4, 5, 6, 3, 2, 3, 2])
print get_indexes('f', 'xsfhhttytffsafweef')
Simply you can go with
a = [['hand', 'head'], ['phone', 'wallet'], ['lost', 'stock']]
b = ['phone', 'lost']
res = [[x[0] for x in a].index(y) for y in b]
Another option
>>> a = ['red', 'blue', 'green', 'red']
>>> b = 'red'
>>> offset = 0;
>>> indices = list()
>>> for i in range(a.count(b)):
... indices.append(a.index(b,offset))
... offset = indices[-1]+1
...
>>> indices
[0, 3]
>>>
And now, for something completely different...
... like confirming the existence of the item before getting the index. The nice thing about this approach is the function always returns a list of indices -- even if it is an empty list. It works with strings as well.
def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
retval = []
last = 0
while val in l[last:]:
i = l[last:].index(val)
retval.append(last + i)
last += i + 1
return retval
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
When pasted into an interactive python window:
Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(the_list, val):
... """Always returns a list containing the indices of val in the_list"""
... retval = []
... last = 0
... while val in the_list[last:]:
... i = the_list[last:].index(val)
... retval.append(last + i)
... last += i + 1
... return retval
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
Update
After another year of heads-down python development, I'm a bit embarrassed by my original answer, so to set the record straight, one can certainly use the above code; however, the much more idiomatic way to get the same behavior would be to use list comprehension, along with the enumerate() function.
Something like this:
def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
return [index for index, value in enumerate(l) if value == val]
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
Which, when pasted into an interactive python window yields:
Python 2.7.14 |Anaconda, Inc.| (default, Dec 7 2017, 11:07:58)
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(l, val):
... """Always returns a list containing the indices of val in the_list"""
... return [index for index, value in enumerate(l) if value == val]
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
And now, after reviewing this question and all the answers, I realize that this is exactly what FMc suggested in his earlier answer. At the time I originally answered this question, I didn't even see that answer, because I didn't understand it. I hope that my somewhat more verbose example will aid understanding.
If the single line of code above still doesn't make sense to you, I highly recommend you Google 'python list comprehension' and take a few minutes to familiarize yourself. It's just one of the many powerful features that make it a joy to use Python to develop code.
Here's a two-liner using Python's index() function:
LIST = ['foo' ,'boo', 'shoo']
print(LIST.index('boo'))
Output: 1
A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry:
>>> a = ['foo','bar','baz','bar','any', 'foo', 'much']
>>> l = dict(zip(set(a), map(lambda y: [i for i,z in enumerate(a) if z is y ], set(a))))
>>> l['foo']
[0, 5]
>>> l ['much']
[6]
>>> l
{'baz': [2], 'foo': [0, 5], 'bar': [1, 3], 'any': [4], 'much': [6]}
>>>
You could also use this as a one liner to get all indices for a single entry. There are no guarantees for efficiency, though I did use set(a) to reduce the number of times the lambda is called.
Finding index of item x in list L:
idx = L.index(x) if (x in L) else -1
This solution is not as powerful as others, but if you're a beginner and only know about forloops it's still possible to find the first index of an item while avoiding the ValueError:
def find_element(p,t):
i = 0
for e in p:
if e == t:
return i
else:
i +=1
return -1
There is a chance that that value may not be present so to avoid this ValueError, we can check if that actually exists in the list .
list = ["foo", "bar", "baz"]
item_to_find = "foo"
if item_to_find in list:
index = list.index(item_to_find)
print("Index of the item is " + str(index))
else:
print("That word does not exist")
List comprehension would be the best option to acquire a compact implementation in finding the index of an item in a list.
a_list = ["a", "b", "a"]
print([index for (index , item) in enumerate(a_list) if item == "a"])
It just uses the python function array.index() and with a simple Try / Except it returns the position of the record if it is found in the list and return -1 if it is not found in the list (like on JavaScript with the function indexOf()).
fruits = ['apple', 'banana', 'cherry']
try:
pos = fruits.index("mango")
except:
pos = -1
In this case "mango" is not present in the list fruits so the pos variable is -1, if I had searched for "cherry" the pos variable would be 2.
There is a more functional answer to this.
list(filter(lambda x: x[1]=="bar",enumerate(["foo", "bar", "baz", "bar", "baz", "bar", "a", "b", "c"])))
More generic form:
def get_index_of(lst, element):
return list(map(lambda x: x[0],\
(list(filter(lambda x: x[1]==element, enumerate(lst))))))
For one comparable
# Throws ValueError if nothing is found
some_list = ['foo', 'bar', 'baz'].index('baz')
# some_list == 2
Custom predicate
some_list = [item1, item2, item3]
# Throws StopIteration if nothing is found
# *unless* you provide a second parameter to `next`
index_of_value_you_like = next(
i for i, item in enumerate(some_list)
if item.matches_your_criteria())
Finding index of all items by predicate
index_of_staff_members = [
i for i, user in enumerate(users)
if user.is_staff()]
Python index() method throws an error if the item was not found. So instead you can make it similar to the indexOf() function of JavaScript which returns -1 if the item was not found:
try:
index = array.index('search_keyword')
except ValueError:
index = -1
name ="bar"
list = [["foo", 1], ["bar", 2], ["baz", 3]]
new_list=[]
for item in list:
new_list.append(item[0])
print(new_list)
try:
location= new_list.index(name)
except:
location=-1
print (location)
This accounts for if the string is not in the list too, if it isn't in the list then location = -1
If you are going to find an index once then using "index" method is fine. However, if you are going to search your data more than once then I recommend using bisect module. Keep in mind that using bisect module data must be sorted. So you sort data once and then you can use bisect.
Using bisect module on my machine is about 20 times faster than using index method.
Here is an example of code using Python 3.8 and above syntax:
import bisect
from timeit import timeit
def bisect_search(container, value):
return (
index
if (index := bisect.bisect_left(container, value)) < len(container)
and container[index] == value else -1
)
data = list(range(1000))
# value to search
value = 666
# times to test
ttt = 1000
t1 = timeit(lambda: data.index(value), number=ttt)
t2 = timeit(lambda: bisect_search(data, value), number=ttt)
print(f"{t1=:.4f}, {t2=:.4f}, diffs {t1/t2=:.2f}")
Output:
t1=0.0400, t2=0.0020, diffs t1/t2=19.60
For those coming from another language like me, maybe with a simple loop it's easier to understand and use it:
mylist = ["foo", "bar", "baz", "bar"]
newlist = enumerate(mylist)
for index, item in newlist:
if item == "bar":
print(index, item)
I am thankful for So what exactly does enumerate do?. That helped me to understand.
Since Python lists are zero-based, we can use the zip built-in function as follows:
>>> [i for i,j in zip(range(len(haystack)), haystack) if j == 'needle' ]
where "haystack" is the list in question and "needle" is the item to look for.
(Note: Here we are iterating using i to get the indexes, but if we need rather to focus on the items we can switch to j.)
I'm trying to figure out how to delete duplicates from 2D list. Let's say for example:
x= [[1,2], [3,2]]
I want the result:
[1, 2, 3]
in this order.
Actually I don't understand why my code doesn't do that :
def removeDuplicates(listNumbers):
finalList=[]
finalList=[number for numbers in listNumbers for number in numbers if number not in finalList]
return finalList
If I should write it in nested for-loop form it'd look same
def removeDuplicates(listNumbers):
finalList=[]
for numbers in listNumbers:
for number in numbers:
if number not in finalList:
finalList.append(number)
return finalList
"Problem" is that this code runs perfectly. Second problem is that order is important. Thanks
finalList is always an empty list on your list-comprehension even though you think it's appending during that to it, which is not the same exact case as the second code (double for loop).
What I would do instead, is use set:
>>> set(i for sub_l in x for i in sub_l)
{1, 2, 3}
EDIT:
Otherway, if order matters and approaching your try:
>>> final_list = []
>>> x_flat = [i for sub_l in x for i in sub_l]
>>> list(filter(lambda x: f.append(x) if x not in final_list else None, x_flat))
[] #useless list thrown away and consumesn memory
>>> f
[1, 2, 3]
Or
>>> list(map(lambda x: final_list.append(x) if x not in final_list else None, x_flat))
[None, None, None, None] #useless list thrown away and consumesn memory
>>> f
[1, 2, 3]
EDIT2:
As mentioned by timgeb, obviously the map & filter will throw away lists that are at the end useless and worse than that, they consume memory. So, I would go with the nested for loop as you did in your last code example, but if you want it with the list comprehension approach than:
>>> x_flat = [i for sub_l in x for i in sub_l]
>>> final_list = []
>>> for number in x_flat:
if number not in final_list:
finalList.append(number)
The expression on the right-hand-side is evalueated first, before assigning the result of this list comprehension to the finalList.
Whereas in your second approach you write to this list all the time between the iterations. That's the difference.
That may be similar to the considerations why the manuals warn about unexpected behaviour when writing to the iterated iterable inside a for loop.
you could use the built-in set()-method to remove duplicates (you have to do flatten() on your list before)
You declare finalList as the empty list first, so
if number not in finalList
will be False all the time.
The right hand side of your comprehension will be evaluated before the assignment takes place.
Iterate over the iterator chain.from_iterable gives you and remove duplicates in the usual way:
>>> from itertools import chain
>>> x=[[1,2],[3,2]]
>>>
>>> seen = set()
>>> result = []
>>> for item in chain.from_iterable(x):
... if item not in seen:
... result.append(item)
... seen.add(item)
...
>>> result
[1, 2, 3]
Further reading: How do you remove duplicates from a list in Python whilst preserving order?
edit:
You don't need the import to flatten the list, you could just use the generator
(item for sublist in x for item in sublist)
instead of chain.from_iterable(x).
There is no way in Python to refer to the current comprehesion. In fact, if you remove the line finalList=[], which does nothing, you would get an error.
You can do it in two steps:
finalList = [number for numbers in listNumbers for number in numbers]
finalList = list(set(finalList))
or if you want a one-liner:
finalList = list(set(number for numbers in listNumbers for number in numbers))
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Accessing the index in Python for loops
I wonder: does Python have something like?
for (i=0; i < length; i += 1){ .... }
Of course, I might say
i = 0
for item in items:
#.....
i += 1
but I think there should be something similar to for(i = 0;...), shouldn't it?
Use the enumerate() function:
for i, item in enumerate(items):
print i, item
or use range():
for i in range(len(items)):
print i
(On python 2 you'd use xrange() instead).
range() let's you step through i in steps other than 1 as well:
>>> list(range(0, 5, 2))
[0, 2, 4]
>>> list(range(4, -1, -1))
[4, 3, 2, 1, 0]
You usually don't need to use an index for sequences though; not with the itertools library or the reversed() function; most usecases for 'special' index value ranges are covered:
>>> menu = ['spam', 'ham', 'eggs', 'bacon', 'sausage', 'onions']
>>> # Reversed sequence
>>> for dish in reversed(menu):
... print(dish)
...
onions
sausage
bacon
eggs
ham
spam
>>> import itertools
>>> # Only every third
>>> for dish in itertools.islice(menu, None, None, 3):
... print(dish)
...
spam
bacon
>>> # In groups of 4
>>> for dish in itertools.izip_longest(*([iter(menu)] * 4)):
... print(dish)
...
('spam', 'ham', 'eggs', 'bacon')
('sausage', 'onions', None, None)
Attempt for that this is a deliberate language choice. The majority of times does loop through numbers in lower level languages, those numbers are used as indexes in a sequence. For the cases one does actually needs the numbers, one should use the range built-in function - which returns an iterable with the desired range([start], end, [step]) input values.
More often, one does need both the items of a sequence and their index, in this case, one should use the builtn enumerateinstead:
for index, letter in enumerate("word"):
print index, letter
And finally, in C and derived languages (Java, PHP, Javascript, C#, Objective C), the for Syntax is just some (very rough) syntax sugar to a while loop - and you can just write the same while loop in Python.
Instead of:
for (start_expr, check_expr, incr_expr) { code_body} you do:
start_expr
while check_expr:
code_body
incr_expr
It works exactly the same.
for i in range(length):
#do something
What range is.