Zha*_*g18 11 max min dataframe pandas elementwise-operations
我有一个Dataframe:
df =
A B C D
DATA_DATE
20170103 5.0 3.0 NaN NaN
20170104 NaN NaN NaN 1.0
20170105 1.0 NaN 2.0 3.0
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我有一个系列
s =
DATA_DATE
20170103 4.0
20170104 0.0
20170105 2.2
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我想运行一个逐元素的max()
函数并s
沿着列的对齐df
.换句话说,我想得到
result =
A B C D
DATA_DATE
20170103 5.0 4.0 NaN NaN
20170104 NaN NaN NaN 1.0
20170105 2.2 NaN 2.2 3.0
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做这个的最好方式是什么?我检查了单列比较和系列到系列的比较,但没有找到一种有效的方法来对一系列运行数据帧.
奖励:不知道答案是不言而喻的上面,但如何做到这一点,如果我要对齐s
沿行的df
(假设尺寸匹配)?
数据:
In [135]: df
Out[135]:
A B C D
DATA_DATE
20170103 5.0 3.0 NaN NaN
20170104 NaN NaN NaN 1.0
20170105 1.0 NaN 2.0 3.0
In [136]: s
Out[136]:
20170103 4.0
20170104 0.0
20170105 2.2
Name: DATA_DATE, dtype: float64
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解:
In [66]: df.clip_lower(s, axis=0)
C:\Users\Max\Anaconda4\lib\site-packages\pandas\core\ops.py:1247: RuntimeWarning: invalid value encountered in greater_equal
result = op(x, y)
Out[66]:
A B C D
DATA_DATE
20170103 5.0 4.0 NaN NaN
20170104 NaN NaN NaN 1.0
20170105 2.2 NaN 2.2 3.0
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我们可以使用以下hack来消除RuntimeWarning
:
In [134]: df.fillna(np.inf).clip_lower(s, axis=0).replace(np.inf, np.nan)
Out[134]:
A B C D
DATA_DATE
20170103 5.0 4.0 NaN NaN
20170104 NaN NaN NaN 1.0
20170105 2.2 NaN 2.2 3.0
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这称为广播,可以按如下方式进行:
import numpy as np
np.maximum(df, s[:, None])
Out:
A B C D
DATA_DATE
20170103 5.0 4.0 NaN NaN
20170104 NaN NaN NaN 1.0
20170105 2.2 NaN 2.2 3.0
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在这里,s[:, None]
将添加一个新轴s
.同样可以通过s[:, np.newaxis]
.当你这样做时,它们可以一起广播,因为形状(3, 4)
和(3, 1)
共同元素.
需要注意的区别s
和s[:, None]
:
s.values
Out: array([ 4. , 0. , 2.2])
s[:, None]
Out:
array([[ 4. ],
[ 0. ],
[ 2.2]])
s.shape
Out: (3,)
s[:, None].shape
Out: (3, 1)
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另一种选择是:
df.mask(df.le(s, axis=0), s, axis=0)
Out:
A B C D
DATA_DATE
20170103 5.0 4.0 NaN NaN
20170104 NaN NaN NaN 1.0
20170105 2.2 NaN 2.2 3.0
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这包括:比较df和s.df较大时,使用df,否则使用s.