How to compare a list in python with another list and replace it with Nan?

Question

delta_list = ['2','3','4',nan]
lb = ['1','2','3','4']
ub = ['5','6','7','8']

condition:

flag_1 = [np.where((np.array(delta_list) > np.array(lb)) & (np.array(delta_list) < np.array(ub)),0,1)]

here: if nan is being compared with any value it is returning 1 (obvious) but I want it to be like if the comparison is with nan in the delta_list , it should always return nan instead of the flag_1 condition . How do I achieve this? Should I make a new list and compare the index with delta_list and replace that index value with nan? Please help!!

Expected Output : flag_1 = ['1','1','1','1'] Desired Output : flag_1 =['1','1','1',nan]

nan because delta_list 4th element is nan and hence

[np.where((np.array(delta_list) > np.array(lb)) & (np.array(delta_list) < np.array(ub)),0,1)] 

should NOT BE evaluated at all if nan is encountered in delta_list and it should simply return nan in my flag_1 list

Answer 1

If numbers are less like 10 is possible compare strings, so use numpy.select with new condition for compare by NaN s:

delta_list = ['2','3','4',np.nan]
lb = ['1','2','3','4']
ub = ['5','6','7','8']

m1 = (np.array(delta_list) > np.array(lb)) & (np.array(delta_list) < np.array(ub))
m2 = pd.isna(delta_list)

flag_1 = np.select([m2, m1], [np.nan, '1'], default='0')
print (flag_1)
['1' '1' '1' 'nan']

If need avoid string nan is possible use None :

flag_1 = np.select([m2, m1], [None, '1'], default='0')
print (flag_1)
['1' '1' '1' None]