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I have a numerical list:
myList = [1, 2, 3, 100, 5]
Now if I sort this list to obtain [1, 2, 3, 5, 100].
What I want is the indices of the elements from the
original list in the sorted order i.e. [0, 1, 2, 4, 3]
--- ala MATLAB's sort function that returns both
values and indices.
If you are using numpy, you have the argsort() function available:
>>> import numpy
>>> numpy.argsort(myList)
array([0, 1, 2, 4, 3])
http://docs.scipy.org/doc/numpy/reference/generated/numpy.argsort.html
This returns the arguments that would sort the array or list.
Something like next:
>>> myList = [1, 2, 3, 100, 5]
>>> [i[0] for i in sorted(enumerate(myList), key=lambda x:x[1])]
[0, 1, 2, 4, 3]
enumerate(myList) gives you a list containing tuples of (index, value):
[(0, 1), (1, 2), (2, 3), (3, 100), (4, 5)]
You sort the list by passing it to sorted and specifying a function to extract the sort key (the second element of each tuple; that's what the lambda is for. Finally, the original index of each sorted element is extracted using the [i[0] for i in ...] list comprehension.
myList = [1, 2, 3, 100, 5]
sorted(range(len(myList)),key=myList.__getitem__)
[0, 1, 2, 4, 3]
I did a quick performance check on these with perfplot (a project of mine) and found that it's hard to recommend anything else but
np.argsort(x)
(note the log scale):
Code to reproduce the plot:
import perfplot
import numpy as np
def sorted_enumerate(seq):
return [i for (v, i) in sorted((v, i) for (i, v) in enumerate(seq))]
def sorted_enumerate_key(seq):
return [x for x, y in sorted(enumerate(seq), key=lambda x: x[1])]
def sorted_range(seq):
return sorted(range(len(seq)), key=seq.__getitem__)
b = perfplot.bench(
setup=np.random.rand,
kernels=[sorted_enumerate, sorted_enumerate_key, sorted_range, np.argsort],
n_range=[2 ** k for k in range(15)],
xlabel="len(x)",
)
b.save("out.png")
The answers with enumerate are nice, but I personally don't like the lambda used to sort by the value. The following just reverses the index and the value, and sorts that. So it'll first sort by value, then by index.
sorted((e,i) for i,e in enumerate(myList))
Updated answer with enumerate and itemgetter:
sorted(enumerate(a), key=lambda x: x[1])
# [(0, 1), (1, 2), (2, 3), (4, 5), (3, 100)]
Zip the lists together: The first element in the tuple will the index, the second is the value (then sort it using the second value of the tuple x[1], x is the tuple)
Or using itemgetter from the operatormodule`:
from operator import itemgetter
sorted(enumerate(a), key=itemgetter(1))
Essentially you need to do an argsort, what implementation you need depends if you want to use external libraries (e.g. NumPy) or if you want to stay pure-Python without dependencies.
The question you need to ask yourself is: Do you want the
indices that would sort the array/list
indices that the elements would have in the sorted array/list
Unfortunately the example in the question doesn't make it clear what is desired because both will give the same result:
>>> arr = np.array([1, 2, 3, 100, 5])
>>> np.argsort(np.argsort(arr))
array([0, 1, 2, 4, 3], dtype=int64)
>>> np.argsort(arr)
array([0, 1, 2, 4, 3], dtype=int64)
Choosing the argsort implementation
If you have NumPy at your disposal you can simply use the function numpy.argsort or method numpy.ndarray.argsort.
An implementation without NumPy was mentioned in some other answers already, so I'll just recap the fastest solution according to the benchmark answer here
def argsort(l):
return sorted(range(len(l)), key=l.__getitem__)
Getting the indices that would sort the array/list
To get the indices that would sort the array/list you can simply call argsort on the array or list. I'm using the NumPy versions here but the Python implementation should give the same results
>>> arr = np.array([3, 1, 2, 4])
>>> np.argsort(arr)
array([1, 2, 0, 3], dtype=int64)
The result contains the indices that are needed to get the sorted array.
Since the sorted array would be [1, 2, 3, 4] the argsorted array contains the indices of these elements in the original.
The smallest value is 1 and it is at index 1 in the original so the first element of the result is 1.
The 2 is at index 2 in the original so the second element of the result is 2.
The 3 is at index 0 in the original so the third element of the result is 0.
The largest value 4 and it is at index 3 in the original so the last element of the result is 3.
Getting the indices that the elements would have in the sorted array/list
In this case you would need to apply argsort twice:
>>> arr = np.array([3, 1, 2, 4])
>>> np.argsort(np.argsort(arr))
array([2, 0, 1, 3], dtype=int64)
In this case :
the first element of the original is 3, which is the third largest value so it would have index 2 in the sorted array/list so the first element is 2.
the second element of the original is 1, which is the smallest value so it would have index 0 in the sorted array/list so the second element is 0.
the third element of the original is 2, which is the second-smallest value so it would have index 1 in the sorted array/list so the third element is 1.
the fourth element of the original is 4 which is the largest value so it would have index 3 in the sorted array/list so the last element is 3.
If you do not want to use numpy,
sorted(range(len(seq)), key=seq.__getitem__)
is fastest, as demonstrated here.
The other answers are WRONG.
Running argsort once is not the solution.
For example, the following code:
import numpy as np
x = [3,1,2]
np.argsort(x)
yields array([1, 2, 0], dtype=int64) which is not what we want.
The answer should be to run argsort twice:
import numpy as np
x = [3,1,2]
np.argsort(np.argsort(x))
gives array([2, 0, 1], dtype=int64) as expected.
Most easiest way you can use Numpy Packages for that purpose:
import numpy
s = numpy.array([2, 3, 1, 4, 5])
sort_index = numpy.argsort(s)
print(sort_index)
But If you want that you code should use baisc python code:
s = [2, 3, 1, 4, 5]
li=[]
for i in range(len(s)):
li.append([s[i],i])
li.sort()
sort_index = []
for x in li:
sort_index.append(x[1])
print(sort_index)
We will create another array of indexes from 0 to n-1
Then zip this to the original array and then sort it on the basis of the original values
ar = [1,2,3,4,5]
new_ar = list(zip(ar,[i for i in range(len(ar))]))
new_ar.sort()
`
s = [2, 3, 1, 4, 5]
print([sorted(s, reverse=False).index(val) for val in s])
For a list with duplicate elements, it will return the rank without ties, e.g.
s = [2, 2, 1, 4, 5]
print([sorted(s, reverse=False).index(val) for val in s])
returns
[1, 1, 0, 3, 4]
Import numpy as np
FOR INDEX
S=[11,2,44,55,66,0,10,3,33]
r=np.argsort(S)
[output]=array([5, 1, 7, 6, 0, 8, 2, 3, 4])
argsort Returns the indices of S in sorted order
FOR VALUE
np.sort(S)
[output]=array([ 0, 2, 3, 10, 11, 33, 44, 55, 66])
Code:
s = [2, 3, 1, 4, 5]
li = []
for i in range(len(s)):
li.append([s[i], i])
li.sort()
sort_index = []
for x in li:
sort_index.append(x[1])
print(sort_index)
Try this, It worked for me cheers!
firstly convert your list to this:
myList = [1, 2, 3, 100, 5]
add a index to your list's item
myList = [[0, 1], [1, 2], [2, 3], [3, 100], [4, 5]]
next :
sorted(myList, key=lambda k:k[1])
result:
[[0, 1], [1, 2], [2, 3], [4, 5], [3, 100]]
A variant on RustyRob's answer (which is already the most performant pure Python solution) that may be superior when the collection you're sorting either:
Isn't a sequence (e.g. it's a set, and there's a legitimate reason to want the indices corresponding to how far an iterator must be advanced to reach the item), or
Is a sequence without O(1) indexing (among Python's included batteries, collections.deque is a notable example of this)
Case #1 is unlikely to be useful, but case #2 is more likely to be meaningful. In either case, you have two choices:
Convert to a list/tuple and use the converted version, or
Use a trick to assign keys based on iteration order
This answer provides the solution to #2. Note that it's not guaranteed to work by the language standard; the language says each key will be computed once, but not the order they will be computed in. On every version of CPython, the reference interpreter, to date, it's precomputed in order from beginning to end, so this works, but be aware it's not guaranteed. In any event, the code is:
sizediterable = ...
sorted_indices = sorted(range(len(sizediterable)), key=lambda _, it=iter(sizediterable): next(it))
All that does is provide a key function that ignores the value it's given (an index) and instead provides the next item from an iterator preconstructed from the original container (cached as a defaulted argument to allow it to function as a one-liner). As a result, for something like a large collections.deque, where using its .__getitem__ involves O(n) work (and therefore computing all the keys would involve O(n²) work), sequential iteration remains O(1), so generating the keys remains just O(n).
If you need something guaranteed to work by the language standard, using built-in types, Roman's solution will have the same algorithmic efficiency as this solution (as neither of them rely on the algorithmic efficiency of indexing the original container).
To be clear, for the suggested use case with collections.deque, the deque would have to be quite large for this to matter; deques have a fairly large constant divisor for indexing, so only truly huge ones would have an issue. Of course, by the same token, the cost of sorting is pretty minimal if the inputs are small/cheap to compare, so if your inputs are large enough that efficient sorting matters, they're large enough for efficient indexing to matter too.
How do I add elements of lists within a list component wise?
p=[[1,2,3],[1,0,-1]]
I have tried the following:
list(map(sum,zip(p[0],p[1])))
Will get me [2,2,2] which is what I need. But how to extend it for a variable number of lists? For example, p=[[1,2,3],[1,0,-1],[1,1,1]] should yield [3,3,3].
A solution I figured out is the following:
import pandas as pd
p=[[1,2,3],[1,0,-1],[1,1,1]]
list(pd.DataFrame(p).sum())
Is there a more "Pythonic" way to solve this problem?
Use * for unpack lists:
a = list(map(sum,zip(*p)))
print (a)
[3, 3, 3]
In numpy solution is similar like in pandas:
a = np.array(p).sum(axis=0).tolist()
print(a)
[3, 3, 3]
You can use * to unpack the list and sum to sum it up.
If you are uncomfortable with the map function you can do it like this:
p = [[1, 2, 3], [4, 5, 6], [-5,-7,-9]]
sum_list = [sum(elem) for elem in zip(*p)]
print(sum_list)
I have a nested list in the following form
inputlist = [[1,2,3],[4,5,6],[7,8,9],[1,2,3,4],[5,6,7,8],[1,2],[3,4]]
I would like further nest it based on changing length as follows:
outputlist = [[[1,2,3],[4,5,6],[7,8,9]],[[1,2,3,4],[5,6,7,8]],[[1,2],[3,4]]]
The underlying logic is that I wish to group every change in list length into a new sublist. It is kind of difficult to explain but I hope the above two examples show what I am trying to do.
How can I achieve this simply and elegantly using python? Thanks.
>>> from itertools import groupby
>>> input_list = [[1,2,3],[4,5,6],[7,8,9],[1,2,3,4],[5,6,7,8],[1,2],[3,4]]
>>> [list(g) for k, g in groupby(input_list, key=len)]
[[[1, 2, 3], [4, 5, 6], [7, 8, 9]], [[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2], [3, 4]]]
Here's an approach.
Get a list of the lengths involved:
#solen: set of lengths
solen = set([len(subl) for subl in inputlist]) # portable
solen = {len[subl] for subl in inputlist} # post Python 2.6
Then build the list of lists of a particular length:
#losubl: list of sublists, one for each item from solen
losubl = [[subl for subl in inputlist if len(subl) == ulen] for ulen in solen]
As jamylak points out, this solution is less efficient than the one based on itertools (more than one pass, sacrifices some order information). OTOH, it may avoid an import if you don't have other uses for itertools. If the lists you're working with are big and complicated, it's probably worth the extra import to use itertools.
I am having a list of lists :
mat = [[1,2,3],[4,5,6],[1,2,3],[7,8,9],[4,5,6]]
and I want to convert into a set i.e. remove the repeating lists and creating a new list out of it which will only contain the unique lists.
In above case the required answer will be
[[1,2,3],[4,5,6],[7,8,9]]
But when I do set(mat), it gives me error
TypeError: unhashable type: 'list'
Can you please solve my problem. Thanks in advance!
Since the lists are mutable, they cannot be hashed. The best bet is to convert them to a tuple and form a set, like this
>>> mat = [[1,2,3],[4,5,6],[1,2,3],[7,8,9],[4,5,6]]
>>> set(tuple(row) for row in mat)
set([(4, 5, 6), (7, 8, 9), (1, 2, 3)])
We iterate through the mat, one list at a time, convert that to a tuple (which is immutable, so sets are cool with them) and the generator is sent to the set function.
If you want the result as list of lists, you can extend the same, by converting the result of set function call, to lists, like this
>>> [list(item) for item in set(tuple(row) for row in mat)]
[[4, 5, 6], [7, 8, 9], [1, 2, 3]]
Lists are mutable, therefore unhashable. Use tuples instead
In [114]: mat = [[1,2,3],[4,5,6],[1,2,3],[7,8,9],[4,5,6]]
In [115]: mat = [tuple(t) for t in mat]
In [116]: matset = set(mat)
In [117]: matset
Out[117]: {(1, 2, 3), (4, 5, 6), (7, 8, 9)}
In [118]: [list(t) for t in matset]
Out[118]: [[4, 5, 6], [7, 8, 9], [1, 2, 3]]
#thefourtheye's answer clearly depicts the problem you were facing with non-hashable data types and the way to by pass it so that you can create a set and remove duplicates. This should suffice for most of thef problems but, re-reading your question
In above case the required answer will be [[1,2,3],[4,5,6],[7,8,9]].
If the order is important, you need to use OrderedDict
>>> from collections import OrderedDict
>>> map(list, OrderedDict.fromkeys(map(tuple, mat)).keys())
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
List as key for dictionary or set is not valid. Because key should remain constant when we want to access values in dictionary for example. Values can change but keys always remain constant.
So in your case:
mat = [[1,2,3],[4,5,6],[1,2,3],[7,8,9],[4,5,6]]
We need to convert inner list to tuple. It can be done by
map(tuple, mat)
Now tuple can be used as key for set/dictionary because tuple's key cannot be changed.
dict.fromkeys(map(tuple, mat))
Since we need final answer as list of list we need to convert all tuple keys of dictionary as list. We don't care about values so we only read keys from dictionary and convert it to list.
mat = map(list, dict.fromkeys(map(tuple, mat)).keys())
Now mat would look something like this.
mat = [[1,2,3],[4,5,6],[7,8,9]]
In Python 3.8+ dictionary will preserve order for older version, OrderedDict can be used.
I'm working with a mapping from values of a python dictionary into a numpy array like this:
import numpy as np
my_array = np.array([0, 1, 2, 3, 4, 5, 6])
my_dict = {'group_a':my_array[0:3], 'group_b':my_array[3:]}
This offers the values referenced through the dict to reflect any changes made in the full array. I need the size of the groups within the dict to be flexible. However when a group is only a single element, such as:
my_dict2 = {'group_a':my_array[0], 'group_b':my_array[1:]}
...then numpy seems to be returning the element value rather than a pointer. The value in the dict no longer reflects any changes in the array. Is there a way to clarify that I want the pointer even for a single element reference?
There is no way to do this that I know of, probably the easiest workaround is to just have the value in the dictionary be a single-element list like so:
my_dict2 = {'group_a':my_array[0:1], 'group_b':my_array[1:]}
ie,
In [2]: my_array = np.array([0, 1, 2, 3, 4, 5, 6])
In [3]: my_dict2 = {'group_a': my_array[0:1], 'group_b': my_array[1:]}
In [4]: my_dict2
Out[4]: {'group_a': array([0]), 'group_b': array([1, 2, 3, 4, 5, 6])}
In [5]: my_array[0] = 200
In [6]: my_dict2
Out[6]: {'group_a': array([200]), 'group_b': array([1, 2, 3, 4, 5, 6])}