将 Python pandas DataFrame 中的数字格式化为以千或百万为单位的货币
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Format numbers in a Python pandas DataFrame as currency in thousands or millions
提问by thatMeow
I have a dataframe: pd.DataFrame({"Amount":[19000000, 9873200, 823449242]}), and I need to convert the numbers into currency ($) in millions. i.e. $19.00MM, $9.88MM, and $823.45MM.
我有一个数据框：pd.DataFrame({"Amount":[19000000, 9873200, 823449242]})，我需要将数字转换为数百万的货币（$）。即 $19.00MM、$9.88MM 和 $823.45MM。
Does anyone know a quick way to do this?
有谁知道一个快速的方法来做到这一点？
Thanks!
谢谢！
回答by EdChum
I think the following should work:
我认为以下应该有效：
df['($) millions'] = '$' + (df['Amount'].astype(float)/1000000).astype(str) + 'MM'
In [3]:
df['($) millions'] = '$' + (df['Amount'].astype(float)/1000000).astype(str) + 'MM'
df
Out[3]:
Amount ($) millions
0 19000000 .0MM
1 9873200 .8732MM
2 823449242 3.449242MM
if needed you can also round:
如果需要，您还可以round：
In [5]:
df['($) millions'] = '$' + (df['Amount'].astype(float)/1000000).round(2).astype(str) + 'MM'
df
Out[5]:
Amount ($) millions
0 19000000 .0MM
1 9873200 .87MM
2 823449242 3.45MM
Another method is to apply a formaton each value using apply:
另一种方法是format使用apply以下方法对每个值应用 a ：
In [15]:
df['($) millions'] = (df['Amount']/1000000).apply(lambda x: '${:,.2f}MM'.format(x))
df
Out[15]:
Amount ($) millions
0 19000000 .00MM
1 9873200 .87MM
2 823449242 3.45MM
However, I expect the first method to scale better for large datasets, although sometimes list comprehensions are faster when it comes to strings
但是，我希望第一种方法可以更好地扩展大型数据集，尽管有时列表理解在字符串方面会更快
Here is the list comprehension method:
这是列表理解方法：
In [17]:
df['($) millions'] = ['${:,.2f}MM'.format(x) for x in df['Amount']/1000000]
df
Out[17]:
Amount ($) millions
0 19000000 .00MM
1 9873200 .87MM
2 823449242 3.45MM
回答by quapka
This simply divides the values - it does not add the $sign etc. (it's only a matter of changing the lambda function), but Amount is still dtype floatso you can treat it as numbers.
这只是将值分开 - 它不添加$符号等（这只是更改 lambda 函数的问题），但 Amount 仍然是 dtype，float因此您可以将其视为数字。
In [41]: df = pd.DataFrame({"Amount":[19000000, 9873200, 823449242]})
In [42]: df['MillionsAsFloat'] = df.apply(lambda row: row['Amount'] / 1000000, axis=1
...: )
In [43]: df['MillionsAsString'] = df.apply(lambda row: '$' + str(round(row['Amount']
...: / 1000000,2)) + 'MM', axis=1)
In [44]: df
Out[44]:
Amount MillionsAsFloat MillionsAsString
0 19000000 19.000000 .0MM
1 9873200 9.873200 .87MM
2 823449242 823.449242 3.45MM
