pandas 熊猫数据框乘以一个系列
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pandas dataframe multiply with a series
提问by jianpan
What is the best way to multiply all the columns of a Pandas DataFrameby a column vector stored in a Series? I used to do this in Matlab with repmat(), which doesn't exist in Pandas. I can use np.tile(), but it looks ugly to convert the data structure back and forth each time.
将 PandasDataFrame的所有列乘以存储在 a 中的列向量的最佳方法是Series什么？我曾经在 Matlab 中使用repmat()，它在 Pandas 中不存在。我可以使用np.tile()，但每次来回转换数据结构看起来很难看。
Thanks.
谢谢。
回答by Wes McKinney
What's wrong with
怎么了
result = dataframe.mul(series, axis=0)
?
?
回答by spencerlyon2
This can be accomplished quite simply with the DataFrame method apply.
这可以通过 DataFrame 方法非常简单地完成apply。
In[1]: import pandas as pd; import numpy as np
In[2]: df = pd.DataFrame(np.arange(40.).reshape((8, 5)), columns=list('abcde')); df
Out[2]:
a b c d e
0 0 1 2 3 4
1 5 6 7 8 9
2 10 11 12 13 14
3 15 16 17 18 19
4 20 21 22 23 24
5 25 26 27 28 29
6 30 31 32 33 34
7 35 36 37 38 39
In[3]: ser = pd.Series(np.arange(8) * 10); ser
Out[3]:
0 0
1 10
2 20
3 30
4 40
5 50
6 60
7 70
Now that we have our DataFrameand Serieswe need a function to pass to apply.
现在我们有了我们的DataFrame，Series我们需要一个函数来传递给apply.
In[4]: func = lambda x: np.asarray(x) * np.asarray(ser)
We can pass this to df.applyand we are good to go
我们可以把它传递给df.apply我们，我们很高兴去
In[5]: df.apply(func)
Out[5]:
a b c d e
0 0 0 0 0 0
1 50 60 70 80 90
2 200 220 240 260 280
3 450 480 510 540 570
4 800 840 880 920 960
5 1250 1300 1350 1400 1450
6 1800 1860 1920 1980 2040
7 2450 2520 2590 2660 2730
df.applyacts column-wise by default, but it can can also act row-wise by passing axis=1as an argument to apply.
df.apply默认情况下按列操作，但它也可以通过axis=1作为参数传递给apply.
In[6]: ser2 = pd.Series(np.arange(5) *5); ser2
Out[6]:
0 0
1 5
2 10
3 15
4 20
In[7]: func2 = lambda x: np.asarray(x) * np.asarray(ser2)
In[8]: df.apply(func2, axis=1)
Out[8]:
a b c d e
0 0 5 20 45 80
1 0 30 70 120 180
2 0 55 120 195 280
3 0 80 170 270 380
4 0 105 220 345 480
5 0 130 270 420 580
6 0 155 320 495 680
7 0 180 370 570 780
This could be done more concisely by defining the anonymous function inside apply
这可以通过在内部定义匿名函数来更简洁地完成 apply
In[9]: df.apply(lambda x: np.asarray(x) * np.asarray(ser))
Out[9]:
a b c d e
0 0 0 0 0 0
1 50 60 70 80 90
2 200 220 240 260 280
3 450 480 510 540 570
4 800 840 880 920 960
5 1250 1300 1350 1400 1450
6 1800 1860 1920 1980 2040
7 2450 2520 2590 2660 2730
In[10]: df.apply(lambda x: np.asarray(x) * np.asarray(ser2), axis=1)
Out[10]:
a b c d e
0 0 5 20 45 80
1 0 30 70 120 180
2 0 55 120 195 280
3 0 80 170 270 380
4 0 105 220 345 480
5 0 130 270 420 580
6 0 155 320 495 680
7 0 180 370 570 780
回答by Andy Hayden
Why not create your own dataframe tile function:
为什么不创建自己的数据框平铺功能：
def tile_df(df, n, m):
dfn = df.T
for _ in range(1, m):
dfn = dfn.append(df.T, ignore_index=True)
dfm = dfn.T
for _ in range(1, n):
dfm = dfm.append(dfn.T, ignore_index=True)
return dfm
Example:
例子：
df = pandas.DataFrame([[1,2],[3,4]])
tile_df(df, 2, 3)
# 0 1 2 3 4 5
# 0 1 2 1 2 1 2
# 1 3 4 3 4 3 4
# 2 1 2 1 2 1 2
# 3 3 4 3 4 3 4
However, the docsnote: "DataFrame is not intended to be a drop-in replacement for ndarray as its indexing semantics are quite different in places from a matrix."Which presumably should be interpreted as "use numpy if you are doing lots of matrix stuff".
但是，文档指出：“DataFrame 并不打算直接替代 ndarray，因为它的索引语义在某些地方与矩阵完全不同。” 这大概应该被解释为“如果你正在做很多矩阵的东西，请使用 numpy”。
