Extracting List Values from Python List

From the question, if the list of variables b is in the list of a, OK appears.
print([i for i in if i in a]) only matches the value.

 a = ['0', '1'', '2', '3', '4', '5', '6', '7', '8', '9']
b = ['0']
if [i for i in if i in a ]:
    print("OK")
else:
    print ("NG")

If you want to know if b elements are also included in a, convert a and b into a set to see if the former includes the latter as a subset.

 a = ['0', '1'', '2', '3', '4', '5', '6', '7', '8', '9']
b = ['0']
if set(b)<=set(a):
    print("OK")
else:
    print ("NG")

If there are duplicate elements, this method is not available.Alternatively, you may want to count the elements using Counter.

 from collections import Counter

a = ['0', '1'', '2', '4', '5', '6', '7', '8', '9']
b = ['0']

counter_a = Counter(a)
counter_b = Counter(b)

# For each element of `b`, is the element contained in `a` and is the number of element contained in `a` greater than that contained in `b`?
if all(map(lambdak:kin counter_a and counter_a[k]>=counter_b[k], counter_b)):
    print('OK')
else:
    print('NG')

They want to extract matching values, so I think it would be better to use the set operation.This program converts variables a and b into set sets, respectively, and
Using the & operator to find the logical product, we are extracting a common element of zero.

 a = ['0', '1'', '2', '3', '4', '5', '6', '7', '8', '9']
b = ['0']

set_a = set(a)
set_b = set(b)

# logical product commonality extraction
set_a&set_b
# >>{'0'}

In order to deepen my understanding, I expressed it in the image.
Also, I think you should read the reference URL for detailed explanations.

Reference URL:
Python formula set
Python set operation Qiita
Ben Diagram Wikipedia

"Since your question was ""a element included in b, there is also a way to use the ""filter"" function to return the list."
(print('OK'), print('NG') may not be the same in the sense...)

 a = ['0', '1'', '2', '3', '4', '5', '6', '7', '8', '9']
b = ['0']
a_in_b=filter(lambdaa_elem:a_elemin b,a)
print(a_in_b)#output: ['0']