How to use list comprehension with in to compare 1D and a 2D list:

Question

I have the lists:

a_list =[['a','1'],['b','2'],['c','3']...]
b_list=['a','b','c']

To access an element that is letters in a_list we would do a_list[0][0] However if I try to do this a_list[0:3][0] I kept getting the same result as a_list[0][0] . How might I get all three elements of the a_list .

This is problem because then I can't use the following code:

all(x in [['a','1'],['b','2'],['c','3']] for x in ['a','b','c'])

The list I have is a lot larger so it would be nice if I can represent as a slice like this:

all(x in a_list[0:3][0] for x in ['a','b','c'])

But the list instead represents the index of the slice not the index of the list contained in the list.

Answer 1

Use operator.itemgetter :

import operator

all(x in map(operator.itemgetter(0), a_list[0:3]) for x in ['a','b','c'])
# True

Answer 2

As per How to efficiently compare two unordered lists (not sets) in Python? , you can use counter

from collections import Counter
Counter(i[0] for i in a_list[:3]) == Counter(b_list)

Answer 3

How about the following? Iterate through the subset of a_list and test if sub-element 0 is part of b_list .

all(x[0] in b_list for x in a_list[0:3])

Answer 4

checking the presence of each element of b_list in a_list and return True or Fasle depends on whether the element is present or not.

a_list =[['a','1'],['b','2'],['c','3']]
b_list=['a','b','c','d']

a_list_1 = list(map(lambda x:x[0],a_list))

result = list(map(lambda a: a in a_list_1, b_list))

"""
output

[True, True, True, False]

"""

Answer 5

Using list-comprehension :

a_list =[['a','1'],['b','2'],['c','3']]
b_list=['a','b','c']

print(all(x[0] in b_list for x in a_list))

OUTPUT :

True