如何格式化浮点数，以成千上万的点，逗号和点分隔?

问题描述

如何格式化来自熊猫数据框的浮点数，以点和逗号分隔数千个，而不是逗号和点分隔? ?

How can I format a float, from a pandas dataframe, to separate thousands by dots and commas, instead by commas and dots? ?

输入:

112299420.40

实际输出:

112,299,420.40

必填输出:

112.299.420,40

我的代码:

pd.set_option('display.float_format', lambda x: '{:,.2f}'.format(x) if abs(x) < 10000 else '{:,.0f}'.format(x))

如何更改这段代码以产生所需的输出?

How can I change this piece of code to produce the required output?

我试图以直观的方式更改它，但没有成功...

I've tried to change it the intuitive way, but without success...

发件人: '{:,.2f}'

收件人: '{:.,2f}'

推荐答案

在这种情况下，Python 2.7字符串格式设置功能似乎无济于事.但是，您可以使用locale模块并将语言环境设置为使用所需的千位/十进制分隔符的国家/地区(在下面的示例中为巴西葡萄牙语):

Python 2.7 string formatting features don't seem to help in this case. However, you can use the locale module and set your locale to a country/language which uses the thousands/decimal separators you want (in the example below, Brazilian Portuguese):

import locale
locale.setlocale(locale.LC_ALL, 'pt_br.utf-8')
pd.set_option('display.float_format', lambda x: locale.format('%.2f', x, grouping=True))

示例:

In [45]: pd.Series([1111.15, 123456.78])
Out[45]:
0     1.111,15
1   123.456,78
dtype: float64

---

如果您要忽略10000以下数字的小数(从代码中看起来是这样):

---

If you want to ignore decimals for numbers under 10000 (as it looks like from your code):

In [49]: pd.set_option('display.float_format', lambda x: locale.format('%.2f', x, grouping=True) if abs(x)>10000 else locale.format('%.0f', x))

In [50]: s
0         1111
1   123.456,78
dtype: float64

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