50 lines
1.9 KiB
Text
50 lines
1.9 KiB
Text
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F commatize(s, period = 3, sep = ‘,’)
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V m = re:‘(\.[0-9]+|[1-9]([0-9]+)?(\.[0-9]+)?)’.search(s)
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I !m
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R s
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V match = m.group()
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V splits = match.split(‘.’)
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V ip = splits[0]
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I ip.len > period
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V inserted = 0
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L(i) ((ip.len - 1) % period + 1 .< ip.len).step(period)
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ip = ip[0 .< i + inserted]‘’sep‘’ip[i + inserted ..]
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inserted += sep.len
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I splits.len > 1
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V dp = splits[1]
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I dp.len > period
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L(i) ((dp.len - 1) I/ period * period .< period - 1).step(-period)
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dp = dp[0 .< i]‘’sep‘’dp[i..]
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ip ‘’= ‘.’dp
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R s[0 .< m.start()]‘’ip‘’s[m.end()..]
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V tests = [‘123456789.123456789’,
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‘.123456789’,
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‘57256.1D-4’,
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‘pi=3.14159265358979323846264338327950288419716939937510582097494459231’,
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‘The author has two Z$100000000000000 Zimbabwe notes (100 trillion).’,
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‘-in Aus$+1411.8millions’,
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‘===US$0017440 millions=== (in 2000 dollars)’,
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‘123.e8000 is pretty big.’,
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‘The land area of the earth is 57268900(29% of the surface) square miles.’,
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‘Ain't no numbers in this here words, nohow, no way, Jose.’,
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‘James was never known as 0000000007’,
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‘Arthur Eddington wrote: I believe there are ’""
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‘15747724136275002577605653961181555468044717914527116709366231425076185631031296’""
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‘ protons in the universe.’,
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‘ $-140000±100 millions.’,
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‘6/9/1946 was a good year for some.’]
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print(commatize(tests[0], period' 2, sep' ‘*’))
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print(commatize(tests[1], period' 3, sep' ‘-’))
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print(commatize(tests[2], period' 4, sep' ‘__’))
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print(commatize(tests[3], period' 5, sep' ‘ ’))
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print(commatize(tests[4], sep' ‘.’))
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L(test) tests[5..]
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print(commatize(test))
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