How To Compare Two Lists And Return The Number Of Times They Match At Each Index In Python?

I have two lists containing 1's and 0's, e.g. list1 = [1,1,0,1,0,1] list2 = [0,1,0,1,1,0] I want to find the number of times they match at each index. So in this case the output w

Solution 1:

zip the lists, compare the elements, compute the sum.

>>>list1 = [1,1,0,1,0,1]>>>list2 = [0,1,0,1,1,0]>>>sum(a == b for a,b inzip(list1, list2))
3

(Consider using itertools.izip in Python 2 for memory efficiency.)

Solution 2:

Here's a lightning fast numpy answer:

import numpy as np
list1 = np.array([1,1,0,1,0,1])
list2 = np.array([0,1,0,1,1,0])

len(np.where(list1==list2)[0])

The numpy np.where function will return the indexes of all the points in the pair of lists that conform to a function (in this case list1==list2 at indices [1,2,3]) along with a datatype description. In the above case, I strip out the array of indices and count how many there are with len().

Solution 3:

You can use map with operator.eq and sum:

>>>import operator>>>sum(map(operator.eq, list1, list2))

This works because True is interpreted as 1 when summed and False like 0.

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You could also use numpy for this:

>>>import numpy as np>>>np.count_nonzero(np.asarray(list1) == np.asarray(list2))