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Simple solution solution in Clear category for Friendly Number by brubru777
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
"""
Format a number as friendly text, using common suffixes.
"""
i = 1
while i < len(powers) and abs(number) >= base ** i:
i += 1
i -= 1
d = number / (base ** i)
d = f'{d:.{decimals}f}' if decimals else int(d)
return f'{d}{powers[i]}{suffix}'
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert friendly_number(102) == '102', '102'
assert friendly_number(10240) == '10k', '10k'
assert friendly_number(12341234, decimals=1) == '12.3M', '12.3M'
assert friendly_number(12461, decimals=1) == '12.5k', '12.5k'
assert friendly_number(1024000000, base=1024, suffix='iB') == '976MiB', '976MiB'
Aug. 22, 2018
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