pandas 如何将布尔值的数据帧转换为 1 和 np.NaN 的数据帧？
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how to convert dataframe of booleans to dataframe of 1 and np.NaN?
提问by d l
I have a dataframe filled with True and False values, and I'd like to get a dataframe from it with the True replaced with 1 and the False replaced with np.NaN. I've tried using dataframe.replace, but it gave a dataframe filled with all True. Is there a way to do it without using for loops and if's?
我有一个填充了 True 和 False 值的数据框，我想从中获取一个数据框，其中 True 替换为 1，False 替换为 np.NaN。我试过使用 dataframe.replace，但它提供了一个填充所有 True 的数据框。有没有办法在不使用 for 循环和 if 的情况下做到这一点？
Example, this is the dataframe I have, with T for True and F for False (not strings 'T' and 'F'; sorry, could not figure out how to format a nicely spaced table in the wiki):
例如，这是我拥有的数据框，T 代表 True，F 代表 False（不是字符串 'T' 和 'F'；抱歉，无法弄清楚如何在 wiki 中格式化一个间隔良好的表格）：
2008-01-02 16:00:00 T T F
2008-01-03 16:00:00 T T T
2008-01-04 16:00:00 T T F
2008-01-07 16:00:00 T T T
2008-01-08 16:00:00 T T F
2008-01-02 16:00:00 TTF
2008-01-03 16:00:00 TTT
2008-01-04 16:00:00 TTF
2008-01-07 16:00:00 TTT
2008-01-08 16 :00:00 TTF
This is what I would like to change it to:
这是我想将其更改为：
2008-01-02 16:00:00 1 1 np.NaN
2008-01-03 16:00:00 1 1 1
2008-01-04 16:00:00 1 1 np.NaN
2008-01-07 16:00:00 1 1 1
2008-01-08 16:00:00 1 1 np.NaN
2008-01-02 16:00:00 1 1 np.NaN
2008-01-03 16:00:00 1 1 1
2008-01-04 16:00:00 1 1 np.NaN
2008-01-07 16: 00:00 1 1 1
2008-01-08 16:00:00 1 1 np.NaN
These are the lines I tried to replace the True and False, and got a dataframe filled with all True values:
这些是我尝试替换 True 和 False 的行，并得到一个填充所有 True 值的数据框：
df.replace(to_replace=True, value=1, inplace=True, method=None)
df.replace(to_replace=False, value=np.NAN, inplace=True, method=None)
When tried separately, the first line alone does not change anything; the second line converts all the values to True.
单独尝试时，仅第一行不会改变任何内容；第二行将所有值转换为 True。
回答by Zelazny7
applymap()can be used to apply a function to every element of a dataframe
applymap()可用于将函数应用于 a 的每个元素 dataframe
In [1]: df = DataFrame([[True, True, False],[False, False, True]]).T
In [2]: df
Out[2]:
0 1
0 True False
1 True False
2 False True
In [3]: df.applymap(lambda x: 1 if x else np.nan)
Out[3]:
0 1
0 1 NaN
1 1 NaN
2 NaN 1
You can also use a dict:
您还可以使用dict：
In [4]: d = {True:1, False:np.nan}
In [5]: df.applymap(lambda x: d[x])
Out[5]:
0 1
0 1 NaN
1 1 NaN
2 NaN 1
Addressing DSM's comment from below. I misread the OP and assumed the datetime was an index. If it's not an index this worked for me:
从下面解决 DSM 的评论。我误读了 OP 并认为日期时间是一个索引。如果它不是对我有用的索引：
In [6]: df.applymap(lambda x: d.get(x,x))
Out[6]:
0 1 2
0 1 NaN 2012-01-01 00:00:00
1 NaN 1 2012-01-01 00:00:00
回答by Jeff
try this. whereworks because the first use by default nans out the not-found entries (e.g. anything that is not == 'T'), then 2nd replaces the non-found entries with the 1
尝试这个。where有效，因为默认情况nan下第一次使用会排除未找到的条目（例如，任何不是 == 'T' 的内容），然后 2nd 将未找到的条目替换为 1
In [48]: df = pd.DataFrame([ 'T', 'T', 'T', 'F', 'F' ], columns=['value'],index=pd.date_range('20010101',periods=5))
In [49]: df
Out[49]:
value
2001-01-01 T
2001-01-02 T
2001-01-03 T
2001-01-04 F
2001-01-05 F
In [50]: df.where(df=='T').where(df!='T',1)
Out[50]:
value
2001-01-01 1
2001-01-02 1
2001-01-03 1
2001-01-04 NaN
2001-01-05 NaN
