pandas应用函数将多个值返回到pandas dataframe中的行

I have a dataframe with a timeindex and 3 columns containing the coordinates of a 3D vector:

我有一个带有timeindex的数据框和包含3D向量坐标的3列:

                         x             y             z
ts
2014-05-15 10:38         0.120117      0.987305      0.116211
2014-05-15 10:39         0.117188      0.984375      0.122070
2014-05-15 10:40         0.119141      0.987305      0.119141
2014-05-15 10:41         0.116211      0.984375      0.120117
2014-05-15 10:42         0.119141      0.983398      0.118164

I would like to apply a transformation to each row that also returns a vector

我想对每个也返回向量的行应用转换

def myfunc(a, b, c):
    do something
    return e, f, g

but if I do:

但如果我这样做:

df.apply(myfunc, axis=1)

I end up with a Pandas series whose elements are tuples. This is beacause apply will take the result of myfunc without unpacking it. How can I change myfunc so that I obtain a new df with 3 columns?

我最终得到了一个Pandas系列,其元素是元组。这是因为申请将取得myfunc的结果而不解压缩它。如何更改myfunc以便获得包含3列的新df?

Edit:

All solutions below work. The Series solution does allow for column names, the List solution seem to execute faster.

以下所有解决方案均可Series系列解决方案允许列名称,List解决方案似乎执行得更快。

def myfunc1(args):
    e=args[0] + 2*args[1]
    f=args[1]*args[2] +1
    g=args[2] + args[0] * args[1]
    return pd.Series([e,f,g], index=['a', 'b', 'c'])

def myfunc2(args):
    e=args[0] + 2*args[1]
    f=args[1]*args[2] +1
    g=args[2] + args[0] * args[1]
    return [e,f,g]

%timeit df.apply(myfunc1 ,axis=1)

100 loops, best of 3: 4.51 ms per loop

%timeit df.apply(myfunc2 ,axis=1)

100 loops, best of 3: 2.75 ms per loop

4 个解决方案

#1

Just return a list instead of tuple.

只需返回一个列表而不是元组。

In [81]: df
Out[81]: 
                            x         y         z
ts                                               
2014-05-15 10:38:00  0.120117  0.987305  0.116211
2014-05-15 10:39:00  0.117188  0.984375  0.122070
2014-05-15 10:40:00  0.119141  0.987305  0.119141
2014-05-15 10:41:00  0.116211  0.984375  0.120117
2014-05-15 10:42:00  0.119141  0.983398  0.118164

[5 rows x 3 columns]

In [82]: def myfunc(args):
   ....:        e=args[0] + 2*args[1]
   ....:        f=args[1]*args[2] +1
   ....:        g=args[2] + args[0] * args[1]
   ....:        return [e,f,g]
   ....: 

In [83]: df.apply(myfunc ,axis=1)
Out[83]: 
                            x         y         z
ts                                               
2014-05-15 10:38:00  2.094727  1.114736  0.234803
2014-05-15 10:39:00  2.085938  1.120163  0.237427
2014-05-15 10:40:00  2.093751  1.117629  0.236770
2014-05-15 10:41:00  2.084961  1.118240  0.234512
2014-05-15 10:42:00  2.085937  1.116202  0.235327

#2

Return Series and it will put them in a DataFrame.

返回系列,它将把它们放在一个DataFrame中。

def myfunc(a, b, c):
    do something
    return pd.Series([e, f, g])

This has the bonus that you can give labels to each of the resulting columns. If you return a DataFrame it just inserts multiple rows for the group.

这样可以为每个结果列提供标签。如果返回DataFrame,则只为该组插入多行。

#3

Based on the excellent answer by @U2EF1, I've created a handy function that applies a specified function that returns tuples to a dataframe field, and expands the result back to the dataframe.

基于@ U2EF1的优秀答案,我创建了一个方便的函数,它应用一个将元组返回到数据帧字段的指定函数,并将结果扩展回数据帧。

def apply_and_concat(dataframe, field, func, column_names):
    return pd.concat((
        dataframe,
        dataframe[field].apply(
            lambda cell: pd.Series(func(cell), index=column_names))), axis=1)

Usage:

df = pd.DataFrame([1, 2, 3], index=['a', 'b', 'c'], columns=['A'])
print df
   A
a  1
b  2
c  3

def func(x):
    return x*x, x*x*x

print apply_and_concat(df, 'A', func, ['x^2', 'x^3'])

   A  x^2  x^3
a  1    1    1
b  2    4    8
c  3    9   27

Hope it helps someone.

希望它可以帮到某人。

#4

Found a possible solution, by changing myfunc to return an np.array like this:

找到一个可能的解决方案,通过更改myfunc返回如下的np.array:

import numpy as np

def myfunc(a, b, c):
    do something
    return np.array((e, f, g))

any better solution?

更好的解决方案?

#1