CODEWITHSUNDEEP
pythonpython-3.xlist
asyl
asyl

Reputation: 31

Python checking if two lists have same value at same index

so im given two lists

lst1 =[0,1,1,1,0]
lst2 =[0,0,1,1,0]

and i need to see which index in both lists have the value 1 here is my code so far

x = list(zip(lst1,lst2))
for i in range(len(x)):
    flag = 0
    for y in range(len(x[i])):
        if x[i][y] == 1:
            flag +=1
    if flag == 2:
        z = x.index(x[i])
        print(z)

but this prints the index 2 and 2 instead of 2 and 3 can anyone point out the problem here, thanks

Upvotes: 0

Views: 2176

Answers (7)

boot-scootin
boot-scootin

Reputation: 12515

You could also use NumPy's element-wise & operation, as well as argwhere for this:

>>> np.argwhere((np.array(lst1) == 1) & (np.array(lst2) == 1)).ravel()
array([2, 3])

Upvotes: 0

David
David

Reputation: 8298

You could do the following:

lst1 =[0,1,1,1,0]
lst2 =[0,0,1,1,0]
assert len(lst1) == len(lst2)

idx = [i for i in range(len(lst1)) if lst1[i] == 1 and lst2[i] == 1]

print(idx)

Other solution using numpy is:

import numpy as np

lst1 =[0,1,1,1,0]
lst2 =[0,0,1,1,0]
assert len(lst1) == len(lst2)
lst1_ = np.array(lst1)
lst2_ = np.array(lst2)

idx_ = np.intersect1d(np.where(lst1_ == 1)[0],np.where(lst2_ == 1)[0])
print(list(idx_))

Other alternative is to switch the following line:

idx_ = np.intersect1d(np.where(lst1_ == 1)[0],np.where(lst2_ == 1)[0])

By:

idx_ = np.where((lst1_==1)&(lst2_==1))[0]

As stated by @yatu is to use bitwise operations.

Upvotes: 1

Nicolas Iceberg
Nicolas Iceberg

Reputation: 743

Look, simple iterate through both lists. Suppose you know length of both lists, then just do following:

lst1 =[0,1,1,1,0]
lst2 =[0,0,1,1,0]

# as we know length of the both lists, and their length are equal, 
# i'll just write length as 5, but you can have other algorhitm of finding length

list_len = 5

is_there_any_matches = False

for index in range(list_len):

    if lst1[index] == lst2[index]:

       is_there_any_matches = True
       break                           # to exit after first match

Note that this will break cycle after first match, but you can remove break, and count number of matches instead. Also, always take length of smaller list to prevent script from error with list index out of range. Have a nice time!

EDIT

I tried to make it as simple as possible, but you can use generators, or other pythonic tricks to make it shorter and more convenient use.

Upvotes: 0

Vignesh Subbiah
Vignesh Subbiah

Reputation: 11

Many answers available on how to do it better.

With respect to your code snippet, List.index always gives the first matching element in the list. That's the reason we are 2,2

>>> lst1 =[0,1,1,1,0]
>>> lst2 =[0,0,1,1,0]
>>> x = list(zip(lst1,lst2))
>>> x
[(0, 0), (1, 0), (1, 1), (1, 1), (0, 0)]
>>> x.index((1,1))
2
>>> x.index((1,1))
2

Upvotes: 1

ccl
ccl

Reputation: 2368

You could do this with element-wise and:

l = [i for i, _ in enumerate(lst1 and lst2) if _ == 1]

With the and function, only in cases where both lists' elements have a value of 1, will the expression return 1, which seems perfect for your question.

Upvotes: -1

boot-scootin
boot-scootin

Reputation: 12515

Assuming they're of the same length, you could use this:

>>> [i for i, (x, y) in enumerate(zip(lst1, lst2)) if x == y == 1]
[2, 3]

Upvotes: 2

Gabio
Gabio

Reputation: 9494

You can iterate the over the values pairs and their index in one loop:

for idx, (i, j) in enumerate(zip(lst1, lst2)):
    if i == j == 1:
        print(idx)

Upvotes: 0

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