python list checking with nested list

Question

I have two python lists. For example:

a = ['1', '2', '3', '4']
b =['1,2', '3,4', '3,33,34', '44']

I need to compare whether list[0] which is one is in b[0] which is 1, 2 and it has to return the output as 1 if it is present and 0 if it is not present.

The final output should be like:

1 (as 1 is present in 1,2)    
0 (as 2 is not present in 3,4)    
1 (as 3 is present in 3,33,34)    
0 (as 4 is not present in 44)

Please do help me in writing a code for this in python as I'm a beginner in this.

Nouman · Accepted Answer · 2018-10-29 10:00:51Z

1

You can also try the following:

for index, value in enumerate(a):  
    if value in b[index].split(","):  
        print(1)  
    else:  
        print(0)

edited Oct 29, 2018 at 10:00

Nouman

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answered Oct 29, 2018 at 6:43

Akhilesh Pandey

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4 Comments

Yadhu Over a year ago

no bro its checking digit by digit and showing the output. I mean for example '4' in '44' is shown as true instead of false. Thanks for trying to help me bro

Akhilesh Pandey Over a year ago

@Yadhu Thanks I have debugged the code. Could you please intend my answer?

Yadhu Over a year ago

thanks bro this worked well and good and bro how do I make the result tat i got in here again back to list? like ['1', '0', '1' ,'0']

Akhilesh Pandey Over a year ago

You can use a list to append the values instead of printing. Before the for loop begins make an empty list named output= [], and add output.append(1) and output.append(0) in place of print(1) and print(0) respectively. Hope it helps you brother

Rakesh · Accepted Answer · 2018-10-29 06:22:58Z

1

Use zip

Ex:

a = ['1','2','3','4']
b =['1,2', '3,4', '3,33,34', '44']

for i, v in zip(a, b):
    if i in v.split(","):     #Check if element in b 
        print(1)
    else:
        print(0)

Output:

answered Oct 29, 2018 at 6:22

Rakesh

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2 Comments

Yadhu Over a year ago

Thanks bro, Thanks a lot this worked like a pro and jiffy.

Yadhu Over a year ago

Oops. Sorry i dint know that as I'm new here too . Done and thanks once again

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