How to split and combine embedded lines using less code

[pjfarley3]

Below is a test program to illustrate what I want to achieve. It is done using several for loops with intermediate variables.

Is there a way to combine these operations into one or two lines using list comprehension in some way? I have not been able to work out how to do this, so I would appreciate any advice you can provide.

Peter

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# Input lists l1 = ["abc\ndef\nghi\n", "jkl\nmno\npqr\n"] l2 = ["123\n456\n789\n", "987\n654\n321\n"]   # Desired result list lwant = ["abc|123\ndef|456\nghi|789\n", "jkl|987\nmno|654\npqr|321\n"]   print("from:\n", l1, "\n", l2, "\nwant:\n", lwant, "\n")   l1s = [] l2s = [] for ln in range(len(l1)):     l1s.append([])     l1s[ln] = l1[ln].splitlines()     l2s.append([])     l2s[ln] = l2[ln].splitlines() print("after splitlines:\n", l1s,     "len(l1s)=%d,len(l1s[0])=%d\n" % (len(l1s), len(l1s[0])),     l2s, "len(l2s)=%d,len(l2s[0])=%d\n" % (len(l2s), len(l2s[0])))   lcomb = [] for ln in range(len(l1)):     lcomb.append([])     for sn in range(len(l1s[0])):         lcomb[ln].append([])         lcomb[ln][sn] = l1s[ln][sn] + "|" + l2s[ln][sn] + "\n" print("after combining:\n%s" % lcomb, "\n")   ljoin = [] for ln in range(len(lcomb)):     ljoin.append("")     ljoin[ln] = "".join(lcomb[ln]) print("after join:\n", ljoin)

Output:
from:
 ['abc\ndef\nghi\n', 'jkl\nmno\npqr\n']
 ['123\n456\n789\n', '987\n654\n321\n']
want:
 ['abc|123\ndef|456\nghi|789\n', 'jkl|987\nmno|654\npqr|321\n']

after splitlines:
 [['abc', 'def', 'ghi'], ['jkl', 'mno', 'pqr']] len(l1s)=2,len(l1s[0])=3
 [['123', '456', '789'], ['987', '654', '321']] len(l2s)=2,len(l2s[0])=3

after combining:
[['abc|123\n', 'def|456\n', 'ghi|789\n'], ['jkl|987\n', 'mno|654\n', 'pqr|321\n']]

after join:
 ['abc|123\ndef|456\nghi|789\n', 'jkl|987\nmno|654\npqr|321\n']

[buran]

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spam = ["abc\ndef\nghi\n", "jkl\nmno\npqr\n"] eggs = ["123\n456\n789\n", "987\n654\n321\n"]   result = [] for chars, nums in zip(spam, eggs):     element = '\n'.join('|'.join(item) for item in zip(chars.strip().split('\n'), nums.strip().split('\n')))     result.append(f'{element}\n') print(result)

Output:
['abc|123\ndef|456\nghi|789\n', 'jkl|987\nmno|654\npqr|321\n']

of course line 6 is ugly and probably should be split in multiple lines, e.g. lines 5-7 above will become

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for chars, nums in zip(spam, eggs):     chars = chars.strip().split('\n')     nums = nums.strip().split('\n')     element = '\n'.join('|'.join(item) for item in zip(chars, nums))     result.append(f'{element}\n')

[pjfarley3]

Thanks much for the prompt reply. One question: Why strip().split("\n") instead of just splitlines()?

Peter

[buran]

Why strip().split("\n") instead of just splitlines()?

no reason - it's perfectly fine to use splitlines() instead
it was easier to do the steps one by one - strip the trailining new line, split at '\n'

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spam = ["abc\ndef\nghi\n", "jkl\nmno\npqr\n"] eggs = ["123\n456\n789\n", "987\n654\n321\n"]   result = [] for chars, nums in zip(spam, eggs):     element = '\n'.join('|'.join(item) for item in zip(chars.splitlines(), nums.splitlines()))     result.append(f'{element}\n') print(result)

[pjfarley3]

Thanks again, that at least shows me I understand the process.

Is it possible to solve this in python V2.7? As I understand it f-strings are V3.6+ specific, correct?

Peter

[buran]

f-strings are 3.6+
earlier versions, incl. 2.7

1	result.append('{}\n'.format(element))

but you shouldn't be using 2.7/ python2 support has ended

[pjfarley3]

Thank you again for your prompt and precise answers.

I will mark this one solved.

Peter