CODEWITHSUNDEEP
pythonpython-3.x
AvePazuzu
AvePazuzu

Reputation: 115

Get single elements from nested list based on index position indecated of the value in a second list

As last part of a bigger project here is what I am trying to solve:

I have a list of lists of which I need to extract exactlty one element based on the value of a second list.

a = [[6,2,3,9], [10,19,14,11], [27,28,21,24]]

b = [0,2,2]

The values in b indicate the positions of the elements in the sublists. Also, the index in b is the true for the index of elements in list a.

The result I am looking for is:

c = [6, 14, 21]

I have tried many versions of this:

c = [i[j] for i in a for j in b]

But as a result I get a list over all emements of all lists looking like this:

c = [6, 3, 3, 10, 14, 14, 27, 21, 21]

Upvotes: 4

Views: 862

Answers (4)

U13-Forward
U13-Forward

Reputation: 71570

Or:

[v[b[i]] for i,v in enumerate(a)]

Example:

>>> a = [[6,2,3,9], [10,19,14,11], [27,28,21,24]]
>>> b = [0,2,2]
>>> [v[b[i]] for i,v in enumerate(a)]
[6, 14, 21]
>>>

Upvotes: 1

Shweta Valunj
Shweta Valunj

Reputation: 158

You can try the following.

a = [[6,2,3,9], [10,19,14,11], [27,28,21,24]]
b = [0,2,2]
c = []
for i in range(0, len(b)):
    c.append(a[i][b[i]])
print (c)

Upvotes: -1

user2390182
user2390182

Reputation: 73450

By using nested loops, you are combining every element from a with every element from b. What you want is pair-wise iteration, using zip:

c = [x[y] for x, y in zip(a, b)]
# [6, 14, 21]

This is roughly equivalent to:

c = [a[i][b[i]] for i in range(min(len(a), len(b)))]

Upvotes: 3

Prem
Prem

Reputation: 85

Try this:

c = [a[i][b[i]] for i in xrange(len(b))]

Upvotes: -1

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