Finding index of number from another list

Question

I am using Python 3, and I have two lists in my code:

ListA = [53, 1, 17, 4, 13, 2, 17] ListB = [4, 3, 1]

Now, I want to find the index of any number in ListB that is in ListA.

The output in this case should be 1 because:

The first value in ListA which is also in ListB is 1.
The index of value 1 in ListA is 1.

You can use list comprehension as index = [ListB.index(j) for j in ListA if j in ListB][0] — Sheldore
– Sheldore, Commented Oct 3, 2018 at 15:55
Depending on your application, np.in1d() could be useful: docs.scipy.org/doc/numpy-1.15.0/reference/generated/… — berkelem
– berkelem, Commented Oct 3, 2018 at 16:01
Should the output be the index of the common value in ListA or ListB? — Patrick Haugh
– Patrick Haugh, Commented Oct 3, 2018 at 16:02
Thanks for fast responses everyone, i will clarify my question. I want to find index in ListA that matches any number from ListB (4, 3 or 1). In example that i posted, number 1 is in ListB and first in ListA (from given numbers in ListB). Expected output is: 2 / or if ListA doesnt have any numbers from ListB, than the output is expexted to be len(ListA). Hope i didn’t confuse you even more now 😅 — Vitali Kazinski
– Vitali Kazinski, Commented Oct 3, 2018 at 16:16

jpp · Accepted Answer · 2018-10-03 16:32:45Z

In pure Python, you can use a generator comprehension with next and enumerate:

A = [53, 1, 17, 4, 13, 2, 17] B = [4, 3, 1] B_set = set(B) first_lst = next(idx for idx, val in enumerate(A) if val in B_set) # 1

Note we hash values in B via set to optimise lookup cost. Complexity is O(m + n), where m and n are number of elements in A and B respectively. To error handle in case no match is found, you can supply a default argument:

first_list = next((idx for idx, val in enumerate(A) if val in B_set), len(A))

If you are happy to use a 3rd party library, you can use NumPy. No error handling here in case of no match:

import numpy as np A = np.array([53, 1, 17, 4, 13, 2, 17]) B = np.array([4, 3, 1]) first_np = np.where(np.in1d(A, B))[0][0] # 1

+1 for numpy. Very convenient if you already work with arrays, so that you don't have to translate them to sets or lists. Also can be adapted for multiple dimensions arrays
This worked just great, thank you friend very much! 😄 btw i used first solution you suggested

blhsing · Accepted Answer · 2018-10-03 16:41:18Z

If you want better efficiency you can turn ListB into a set so that you can determine if an item is in ListB with an average time complexity of O(1):

setB = set(ListB) print(next(i for i, a in enumerate(ListA) if a in setB))

This outputs: 1

blhsing · Accepted Answer · 2018-10-03 16:42:38Z

1

You can use the following generator expression:

next(i for i, a in enumerate(ListA) for b in ListB if a == b)

Given your sample input, this returns: 1

edited Oct 3, 2018 at 16:42

answered Oct 3, 2018 at 15:59

blhsing

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3 Comments

jpp Over a year ago

Nice, but the nested for loop makes complexity O(m * n), right? [where m, n = len(A), len(B)]. Also index should refer to A, not B.

mad_ Over a year ago

You should enumerate ListA rather than ListB

blhsing Over a year ago

@jpp Yes, I wrote this answer in order to give an answer quickly. ;-) I just posted a much more efficient answer separately. You can check it out if you want. Thanks.

mad_ · Accepted Answer · 2018-10-03 16:38:29Z

Set intersection to find the common values. Then find all the indices which are present in ListA and then find the minimum index. If in case there is no match it will print the length of ListA

set_inter =set(ListB).intersection(ListA) if set_inter: # if there is a common value idx_A=min([ListA.index(i) for i in set_inter]) print(idx_A) else: print(len(ListA)) # print the length of ListA

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