python人类可读的大数[重复]

据我了解，您只需要“最重要的”部分。为此，请使用

floor(log10(abs(n)))

import mathmillnames = ['',' Thousand',' Million',' Billion',' Trillion']def millify(n):    n = float(n)    millidx = max(0,min(len(millnames)-1,  int(math.floor(0 if n == 0 else math.log10(abs(n))/3))))    return '{:.0f}{}'.format(n / 10**(3 * millidx), millnames[millidx])

为多个不同的数字运行以上函数：

for n in (1.23456789 * 10**r for r in range(-2, 19, 1)):    print('%20.1f: %20s' % (n,millify(n)))      0.0:         0      0.1:         0      1.2:         1     12.3:        12    123.5:       123   1234.6:1 Thousand  12345.7:          12 Thousand 123456.8:         123 Thousand1234567.9: 1 Million          12345678.9:12 Million         123456789.0:          123 Million        1234567890.0: 1 Billion       12345678900.0:12 Billion      123456789000.0:          123 Billion     1234567890000.0:1 Trillion    12345678900000.0:          12 Trillion   123456789000000.0:         123 Trillion  1234567890000000.0:        1235 Trillion 12345678899999998.0:       12346 Trillion123456788999999984.0:      123457 Trillion1234567890000000000.0:     1234568 Trillion