CODEWITHSUNDEEP
pythonlistmaxsublist
navajyothmp
navajyothmp

Reputation: 95

Finding out indices of maxima of sublists in python

I've a list x of n sublists. I want to make a new_list of length n containing the indices of maximum of each sublist. How can I pull this off?

For example:

x = [[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]

(I've taken the maxima the same in each sublist for the ease of use)

And the output must hence be:

new_list = [4,1,0,3,2]

Upvotes: 0

Views: 947

Answers (4)

Keerthana Prabhakaran
Keerthana Prabhakaran

Reputation: 3787

You can use lambda expression and map over l, to get the index of max element in x

l=[[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]
print map(lambda x:x.index(max(x)),l)

Output:

[4, 1, 0, 3, 2]

If you are using numpy array,

>>> print map(lambda x:int(np.where(x==max(x))[0]),l)
[4, 1, 0, 3, 2]

Upvotes: 1

fileyfood500
fileyfood500

Reputation: 1321

Make a new list by looping through and getting the max of each sublist:

Like so:

x = [[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]

max_index = [i.index(max(i)) for i in x]

Upvotes: -1

Akavall
Akavall

Reputation: 86178

If you have access to numpy it is very easy:

In [9]: import numpy as np

In [10]: x = [[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]

In [11]: a = np.array(x) # Assumes all sublists are of the same length

In [12]: np.argmax(a, axis=1)
Out[12]: array([4, 1, 0, 3, 2])

Upvotes: 1

A.J. Uppal
A.J. Uppal

Reputation: 19264

Provided there are no duplicate items in each sublist, try the following:

new_list = [sub.index(max(sub)) for sub in x]

>>> x = [[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]
>>> new_list = [sub.index(max(sub)) for sub in x]
>>> new_list
[4, 1, 0, 3, 2]
>>>

Upvotes: 1

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