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Comma quibbling: Difference between revisions
→{{header|Python}}: Second example without the need to reverse a string.
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(→{{header|Python}}: Second example without the need to reverse a string.) |
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=={{header|Python}}==
===replace() whilst reversed===
<lang python>>>> def strcat(sequence):
▲ # replace(..) can only replace the first X occurrences not the last
▲ # Hence the replace is done on the reverse of the intermediate string
▲ # then reversed back.
▲ return '{%s}' % ', '.join(sequence)[::-1].replace(',', 'dna ', 1)[::-1]
>>> for seq in ([], ["ABC"], ["ABC", "DEF"], ["ABC", "DEF", "G", "H"]):
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Input: ['ABC', 'DEF', 'G', 'H'] -> Output: '{ABC, DEF, G and H}'
>>> </lang>
===Counted replacement===
replace() will replace nothing if the count of items to replace is zero, (and negative integer counts act to replace all occurrences).
This combines with the length of the input sequence to allow this to work:
<lang python>def commaQuibble(s):
return '{%s}' % ' and '.join(s).replace(' and ', ', ', len(s) - 2)
for seq in ([], ["ABC"], ["ABC", "DEF"], ["ABC", "DEF", "G", "H"]):
print('Input: %-24r -> Output: %r' % (seq, commaQuibble(seq)))</lang>
{{out}}
<pre>Input: [] -> Output: '{}'
Input: ['ABC'] -> Output: '{ABC}'
Input: ['ABC', 'DEF'] -> Output: '{ABC and DEF}'
Input: ['ABC', 'DEF', 'G', 'H'] -> Output: '{ABC, DEF, G and H}'</pre>
=={{header|PL/I}}==
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