Python - Replace specific element in nested list comprehension

Question

Preface: I realized this is just me being obsessed with making something more pythonic.

I have a list of lists like such:

L = [[1,'',3,''],[1,2,'',4],[1,2,3,''],['',2,3,4]]

I need to replace ONLY the 4th element with the number 4 IF it is ' '.

This can be achieved with a simple for loop:

for row in L:
    if row[3] =='':
        row[3] = 4

How can I achieve this through a nested list comprehension?

My best attempt is the following, but it results in a list of lists that have all values of ' ' replaced with 4, rather than the specific element.

L = [[4 if x=='' else x for x in y] for y in L]

Ajax1234 · Accepted Answer · 2017-06-15 19:22:16Z

2

You can try this:

L = [[1,'',3,''],[1,2,'',4],[1,2,3,''],['',2,3,4]]

new_l = [[4 if b == '' and c == 3 else b for c, b in enumerate(d)] for d in L]

Output:

[[1, '', 3, 4], [1, 2, '', 4], [1, 2, 3, 4], ['', 2, 3, 4]]

By using enumerate, you can determine if both the element itself is equal to '' and verify that the index of the occurrence is 3, which is the fourth element.

answered Jun 15, 2017 at 19:22

Ajax1234

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Comments

zwer · Accepted Answer · 2017-06-15 19:35:58Z

0

This can be done in a very simple manner, and quite performant to boot (executes in O(N) time almost entirely on the C side), if all your sublists are of the same length:

L = [[1, '', 3, ''], [1, 2, '', 4], [1, 2, 3, ''], ['', 2, 3, 4]]

L2 = [[e0, e1, e2, 4 if e3 == "" else e3] for e0, e1, e2, e3 in L]
# [[1, '', 3, 4], [1, 2, '', 4], [1, 2, 3, 4], ['', 2, 3, 4]]

However, Pythonic is an illusive term and means different things to different people - cramming everything into an one-liner is not necessarily 'Pythonic'.

UPDATE - Actually, going on the same track, you can make it sub-list size agnostic, and replace the last element without knowing upfront the number of elements or if they even differ:

L = [[1, '', 3, ''], [1, 2, ''], [1, 2, 3, '', 5], ['', 2, 3, 4, 5, '']]

L2 = L2 = [e[:-1] + ["FOO" if e[-1] == "" else e[-1]] for e in L]
# [[1, '', 3, 'FOO'], [1, 2, 'FOO'], [1, 2, 3, '', 5], ['', 2, 3, 4, 5, 'FOO']]

edited Jun 15, 2017 at 19:35

answered Jun 15, 2017 at 19:26

zwer

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1 Comment

Błotosmętek Over a year ago

This will also replace 0.

Ankan Roy · Accepted Answer · 2017-06-15 19:49:24Z

0

There is no need to put two for loops , this can be achieved with one loop only

def lis(L):
    for y in L:
        if y[3] == '':
            y[3] = 4
print L

lis([[1,'',3,''],[1,2,'',4],[1,2,3,''],['',2,3,4]])

answered Jun 15, 2017 at 19:49

Ankan Roy

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1 Comment

ThousandFacedHero Over a year ago

I believe this is the same for loop solution I put in the OP. I was trying to find a list comprehension to do the same.

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