CODEWITHSUNDEEP
python
Sylvia Chang
Sylvia Chang

Reputation: 29

How to remove elements in a list based on another list in python, without loops?

I have two lists of equal length:

list_a = ['a','b','c','d']
list_b = [-6.3, 3.1, 0.5, 4.1]

I want to remove the elements < 0.7 in list_b, and simultaneously remove the corresponding elements from list_a, i.e.

list_a_2 = ['b','d']
list_b_2 = [3.1, 4.1]

I know the second list, 

list_b_2 = [item for item in hem if item > 0.7].

But is there a list-thinking way to get list_a_2, without using loops?

Upvotes: 2

Views: 88

Answers (4)

Sylvia Chang
Sylvia Chang

Reputation: 29

Thanks a lot! For other people having a same problem:

I end up with the following code:

list_a = ['a','b','c','d']
list_b = [-6.3, 3.1, 0.5, 4.1]
tmp = zip(a, b)
list_a_2 = [x[0] for x in tmp if x[1] > 0.7]
list_b_2 = [x[1] for x in tmp if x[1] > 0.7]
del tmp

And a better idea might be directly use a dataframe to match each element with its own label.

Upvotes: 0

akuiper
akuiper

Reputation: 214927

One way is to use zip:

list_a_2, list_b_2 = zip(*[(a, b) for a, b in zip(list_a, list_b) if b > 0.7])    
list_a_2
# ('b', 'd')

list_b_2
# (3.1, 4.1)

If a for loop is better suited, you can create two empty lists and conditionally append the values to them:

list_a_2, list_b_2 = [], []
for a, b in zip(list_a, list_b):
    if b > 0.7:
        list_a_2.append(a)
        list_b_2.append(b)

list_a_2
# ['b', 'd']

list_b_2
# [3.1, 4.1]

Upvotes: 6

bunji
bunji

Reputation: 5213

If you really want to avoid loops you can use a recursive function:

def unnecessarily_recursive_function(list_a,list_b):
    try:
        a = list_a.pop(0)
        b = list_b.pop(0)
        tmp_a, tmp_b = unnecessarily_recursive_function(list_a,list_b)
        if b < 0.7:
            return [a] + tmp_a, [b] + tmp_b
        else:
            return tmp_a, tmp_b
    except IndexError:
        return [],[]

which you can call like:

list_a, list_b = unnecessarily_recursive_function(list_a, list_b)

Note that this is a really bad reason to use a recursive function and you should definitely actually go with the other suggestions above.

Upvotes: 0

willnx
willnx

Reputation: 1283

Without explicit loops, create both? Sure, if we use a temporary object:

list_a = ['a','b','c','d']
list_b = [-6.3, 3.1, 0.5, 4.1]
tmp = zip(a, b)
list_a_2 = [x[0] for x in tmp if x[1] > 0.7]
list_b_2 = [x[1] for x in tmp if x[1] > 0.7]
del tmp

But using an actual for-loop will be a bit more obvious:

for idx, value in enumerate(list_b):
    if value =< 0.7:
        list_a.pop(idx)
        list_b.pop(idx)

But you're still managing two lists for mapping (basically) keys to values. This is what a dictionary is for! Consolidate those list and create your subset with some dictionary comprehension:

{x:y for x, y in zip(list_a, list_b) if y > 0.7}

Upvotes: 1

Related Questions