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HedgeHog
Asked:2022-07-18 21:11:10 +0000 UTC2022-07-18 21:11:10 +0000 UTC

缩位数字输出

  • 772

不久前我有一个以缩写格式显示大小值的功能，（千，百万等），现在是如何实现的，我想吊死自己：D，这可以通过某种方式改进吗？或者也许有现成的模块？

def textify_value(value):
    avalue = abs(value)
    if avalue >= 1000000000000000000000000000000000000000000000000000000000000000:
        return str(round(value / 1000000000000000000000000000000000000000000000000000000000000000)) + " эндек."
    if avalue >= 1000000000000000000000000000000000:
        return str(round(value / 1000000000000000000000000000000000)) + " дек."
    if avalue >= 1000000000000000000000000000000:
        return str(round(value / 1000000000000000000000000000000)) + " нон."
    if avalue >= 1000000000000000000000000000:
        return str(round(value / 1000000000000000000000000000)) + " окт."

    if avalue >= 1000000000000000000000000:
        return str(round(value / 1000000000000000000000000)) + " сптл."

    if avalue >= 1000000000000000000000:
        return str(round(value / 1000000000000000000000)) + " скст."

    if avalue >= 1000000000000000000:
        return str(round(value / 1000000000000000000)) + " квнт."

    if avalue >= 1000000000000000:
        return str(round(value / 1000000000000000)) + " квдр."

    if avalue >= 1000000000000:
        return str(round(value / 1000000000000)) + " трлн."

    if avalue >= 1000000000:
        return str(round(value / 1000000000)) + " млрд."

    if avalue >= 1000000:
        return str(round(value / 1000000)) + " млн."

    if avalue >= 100000:
        return str(round(value / 100000)) + "00 тыс."

    if avalue >= 1000:
        return str(round(value / 1000)) + " тыс."

    return str(value)
python

3 个回答

  1. 1_000_000 =1e+6

    其余的以此类推。它被称为指数符号。这是为了使 ifs 看起来正常。

    例如，对于单词，您可以使用字典或列表。

    • 1
  2. Best Answer
    def f(value):
    arr = [(0, ''), (3, ' тыс.'), (6, ' млн.'), (9, ' млрд.'), (12, ' трлн.'), (15, ' квдр.'), (18, ' квнт.'), (21, ' скст.'), (24, ' сптл.'), (27, ' окт.'), (30, ' нон.'), (33, ' дек.'), (63, ' эндек.')]
    for n, s in arr[::-1]:
        n = 10**n
        if abs(value) >= n:
            return str(round(value / n)) + s
    return str(value)
    • 1

  3. 是的，使用人性化模块

    import humanize
humanize.i18n.activate("ru_RU")
humanize.clamp(200000000000000000000000000000000, format=humanize.intword) #other param , ceil=3000, floor=1000, floor_token='lower', ceil_token='more'
#200.0 нониллиона

    但他似乎只指望dedocelion

    • 1

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