使用 NaN 向下舍入 Pandas 数据框列中的值
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Rounding down values in Pandas dataframe column with NaNs
提问by user1718097
I have a Pandas dataframe that contains a column of float64 values:
我有一个 Pandas 数据框，其中包含一列 float64 值：
tempDF = pd.DataFrame({ 'id': [12,12,12,12,45,45,45,51,51,51,51,51,51,76,76,76,91,91,91,91],
'measure': [3.2,4.2,6.8,5.6,3.1,4.8,8.8,3.0,1.9,2.1,2.4,3.5,4.2,5.2,4.3,3.6,5.2,7.1,6.5,7.3]})
I want to create a new column containing just the integer part. My first thought was to use .astype(int):
我想创建一个仅包含整数部分的新列。我的第一个想法是使用 .astype(int)：
tempDF['int_measure'] = tempDF['measure'].astype(int)
This works fine but, as an extra complication, the column I have contains a missing value:
这工作正常，但作为一个额外的复杂因素，我的列包含一个缺失值：
tempDF.ix[10,'measure'] = np.nan
This missing value causes the .astype(int) method to fail with:
这个缺失值导致 .astype(int) 方法失败：
ValueError: Cannot convert NA to integer
I thought I could round down the floats in the column of data. However, the .round(0) function will round to the nearest integer (higher or lower) rather than rounding down. I can't find a function equivalent to ".floor()" that will act on a column of a Pandas dataframe.
我以为我可以舍入数据列中的浮点数。但是， .round(0) 函数将舍入到最接近的整数（更高或更低）而不是向下舍入。我找不到相当于“.floor()”的函数，它可以作用于 Pandas 数据框的列。
Any suggestions?
有什么建议？
回答by Joachim Isaksson
You could just apply numpy.floor;
你可以申请numpy.floor;
import numpy as np
tempDF['int_measure'] = tempDF['measure'].apply(np.floor)
id measure int_measure
0 12 3.2 3
1 12 4.2 4
2 12 6.8 6
...
9 51 2.1 2
10 51 NaN NaN
11 51 3.5 3
...
19 91 7.3 7
回答by Alexander
You could also try:
你也可以试试：
df.apply(lambda s: s // 1)
Using np.flooris faster, however.
np.floor但是，使用速度更快。
回答by ledawg
The answers here are pretty dated and as of pandas 0.25.2 (perhaps earlier) the error
这里的答案已经过时了，从 pandas 0.25.2（可能更早）开始，错误
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
Which would be
哪个是
df.iloc[:,0] = df.iloc[:,0].astype(int)
for one particular column.
对于一个特定的列。
