Filter a string based on binary sequence using itertools.compress

Question

I have a string and binary sequence(which is int of 0s and 1s)

sent1 = 'ERAGSJKDLLDERRR'
bin_seq = 100101010100011

So I want to fetch the letters by comparing with bin_seq. So it will return the letter if the value of bin_seq is 1 at corresponding positions.

So it should return:

'EGJDLRR'

I am using itertools.compress for the above operation.

from itertools import compress

sent1 = 'ERAGSJKDLLDERRR'
bin_seq = 100101010100011

print("".join(list(itertools.compress(sent1, str(bin_seq)))))

Which returns the output:

'ERAGSJKDLLDERRR'

I know I can do it easily by using for loop:

sent_new = []
for i,j in zip(sent1, str(bin_seq)):
    if j == '1':
        sent_new.append(i)
print("".join(sent_new))

But I am more concern about why it is not giving expected output with itertools.compress.

You could also compress (pun intended) your zip approach in a single line: "".join(c for c, b in zip(sent1, str(bin_seq)) if b == '1') — tobias_k
– tobias_k, Commented Feb 15, 2019 at 15:07
@tobias_k I am practicing with itertools so I was more concern about itertool.compress solution. — Sociopath
– Sociopath, Commented Feb 15, 2019 at 15:12
BTW, IMHO using an int for bin_seq is not a good choice. You should only use numeric types if you intend to do math, e.g. not for phone numbers, zip codes, or other strings that just happen to be all-numbers. — tobias_k
– tobias_k, Commented Feb 15, 2019 at 15:12
You caused the problem by trying to pass in bin_seq = 100101010100011, which makes it an integer, not a binary sequence (or even a string). The problem then becomes how to convert it back to a sequence (or iterable) of binary digits. So the question title is a misnomer: your "binary sequence" stopped being a binary sequence the moment you converted it to integer, which makes tthings harder. — smci
– smci, Commented Feb 18, 2024 at 2:19

Kevin · Accepted Answer · 2019-02-15 15:09:29Z

2

Your compress approach is close. It's not working because the strings "0" and "1" both evaluate to True in a boolean context. One quick solution is to convert them to ints, since 0 is False and 1 is True in a boolean context:

import itertools

sent1 = 'ERAGSJKDLLDERRR'
bin_seq = 100101010100011

print("".join(itertools.compress(sent1, map(int, str(bin_seq)))))

Result:

EGJDLRR

edited Feb 15, 2019 at 15:09

answered Feb 15, 2019 at 15:03

Kevin

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gold_cy · Accepted Answer · 2019-02-15 15:04:03Z

1

from itertools import compress

''.join(compress(sent1, map(int, str(bin_seq))))

'EGJDLRR'

The issue is you need to feed compress a list of (0, 1) as ints.

answered Feb 15, 2019 at 15:04

gold_cy

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Comments

SpghttCd · Accepted Answer · 2019-02-16 21:37:00Z

1

You could use LC instead of for loop

''.join([c for c, b in zip(sent1, str(bin_seq)) if b=='1'])

Edit:
For the sake of interest I timed the two by now available solutions:

''.join(compress(sent1, map(int, str(bin_seq))))
1.360 s                                                    
''.join([c for c, b in zip(sent1, str(bin_seq)) if b=='1']) 
0.766 s

edited Feb 16, 2019 at 21:37

answered Feb 15, 2019 at 15:13

SpghttCd

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