Iterating through 2 lists and checking for values

Question

I wish to see whether values x exists in values y of a list, if they do, I want to print out the total number of such values.

z=0
for x,y in zip(labels,n):
    if x in y:
        z=z+1
print(z)

This is what labels looks like:

[['20011', '20048'],
 ['20011', '20048'],
 ['20011', '20048'],
 ['20011', '20048']]

And this is what n looks like:

['20011', '20048' ,'20011', '20048']

I get a zero on printing out z. What am I doing wrong? If I don't define z, I get an error saying no z defined.

Do you mean to find wheter the lists in `labels` are contained in `n`? — Martin, Feb 02 '16 at 15:57
No, the other way round.. whether the values in n are contained in labels by iterating row wise. — minks, Feb 02 '16 at 15:59
@jdkc: But then I receive an error that says: *TypeError: 'in ' requires string as left operand, not numpy.ndarray* — minks, Feb 02 '16 at 16:00
@CoderQueen Check my answer, it converts your list into `int`. — Torxed, Feb 02 '16 at 16:02

Padraic Cunningham · Answer 1 · 2016-02-02T17:07:53.953

If you want the count if the elements appear anywhere, get the union of all the sublists and see how many times each element from l2 appears in it:

l = [['20011', '20048'],
 ['20011', '20048'],
 ['20011', '20048'],
 ['20011', '20048']]


l2 = ['20011', '20048' ,'20011', '20048']
union = set.union(*map(set,l))
print(sum(ele in union for ele in l2)) # ->  4

If you don't want to count unique elements more than once, get the intersection:

l = [['20011', '20048'],
     ['20011', '20048'],
     ['20011', '20048'],
     ['20011', '20048']]

l2 = ['20011', '20048', '20011', '20048']
inter = set.union(*map(set, l)).intersection(l2)

print(len(inter)) # ->  2

If you want to use the elements from the sublists for the count:

l = [['20011', '20048'],
     ['20011', '20048'],
     ['20011', '20048'],
     ['20011', '20048']]

l2 = ['20011', '20048', '20011', '20048']
st = set(l2)
from itertools import chain
print(sum(ele in st for ele in chain.from_iterable(l)))

To count based on the sublist being disjoint or not from n, you can use set.isdisjoint so if there is any common elements not st.isdisjoint(sub) will be True.:

l = [['20011', '20048'],
     ['20011', '20048'],
     ['20011', '20048'],
     ['20011', '20048']]

l2 = ['20011', '20048', '20011', '20048']
st = set(l2)
print(sum(not st.isdisjoint(sub) for sub in l)) # -> 4

This is very helpful. Thanks a lot! – minks Feb 02 '16 at 17:07 — minks, Feb 02 '16 at 17:07

cromod · Accepted Answer · 2016-02-02T17:06:45.207

1

numpy.intersect1d enables to get the intersection between 2 arrays, then you can use size to get the count of unique elements in both arrays :

import numpy as np
labels = np.array([[20011,20048],[20011,20048],[20011,20048],[20011,20048]])
n = np.array([20011,20048,20011,20048])
z = np.intersect1d(n,labels).size
print z # counts 2

numpy.in1d enables to check if each elements of an 1D array are in a 2nd array, then you convert into list and count the True item (not unique elements) :

z = np.in1d(n,labels).tolist().count(True)
print z # counts 4

edited Feb 02 '16 at 17:06

answered Feb 02 '16 at 16:05

cromod

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26

This gives me a 0. Is it because labels is a list of list while n is just a list? – minks Feb 02 '16 at 16:08
I get 2 with your example – cromod Feb 02 '16 at 16:17
Hi, this works perfectly. I had some issue in my code. My bad. Thanks a lot! – minks Feb 02 '16 at 16:26
you've numpy.in1d too. It provides an array to check if each element of n is in labels. In this case, you'll count 4. – cromod Feb 02 '16 at 16:44
Why does the difference exist in the z values then? One gives 2 and the other 4? Which one is the right case? – minks Feb 02 '16 at 16:54
It depends on your need. With intersect1d you'll get the number of unique elements that are in both n and labels. With in1d the number of elements n that are in labels. – cromod Feb 02 '16 at 17:01

Iterating through 2 lists and checking for values

2 Answers2