Question

让我们采用这个数据框：

import pandas as pd
L0 = ['d','a','b','c','d','a','b','c','d','a','b','c']
L1 = ['z','z','z','z','x','x','x','x','y','y','y','y']
L2 = [1,6,3,8,7,6,7,6,3,5,6,5]
df = pd.DataFrame({"A":L0,"B":L1,"C":L2})
df = df.pivot(columns="A",index="B",values="C")

旋转后，列和行按字母顺序排列。

重新排序列非常简单，可以使用自定义列标签列表来完成：

df = df[['d','a','b','c']]

但重新排序行没有这样的直接功能，我能想到的最优雅的方法是使用列标签功能并前后移调：

df = df.T[['z','x','y']].T

这样做，例如根本没有效果：

df.loc[['x','y','z'],:] = df.loc[['z','x','y'],:]

通过提供索引标签的自定义列表，没有直接的方法来对数据行的行进行排序吗？

Answer 1

您可以使用reindex或reindex_axis，速度更快loc：

index：

idx = ['z','x','y']
df = df.reindex(idx)
print (df)
A  a  b  c  d
B            
z  6  3  8  1
x  6  7  6  7
y  5  6  5  3

或者：

idx = ['z','x','y']
df = df.reindex_axis(idx)
print (df)
A  a  b  c  d
B            
z  6  3  8  1
x  6  7  6  7
y  5  6  5  3

正如ssm所指出的那样：

df = df.loc[['z', 'x', 'y'], :]
print (df)
A  a  b  c  d
B            
z  6  3  8  1
x  6  7  6  7
y  5  6  5  3

对于列：

cols = ['d','a','b','c']
df = df.reindex(columns=cols)
print (df)
A  d  a  b  c
B            
x  7  6  7  6
y  3  5  6  5
z  1  6  3  8

cols = ['d','a','b','c']
df = df.reindex_axis(cols, axis=1)
print (df)
A  d  a  b  c
B            
x  7  6  7  6
y  3  5  6  5
z  1  6  3  8

这两种：

idx = ['z','x','y']
cols = ['d','a','b','c']
df = df.reindex(columns=cols, index=idx)
print (df)
A  d  a  b  c
B            
z  1  6  3  8
x  7  6  7  6
y  3  5  6  5

<强>计时：

In [43]: %timeit (df.loc[['z', 'x', 'y'], ['d', 'a', 'b', 'c']])
1000 loops, best of 3: 653 µs per loop

In [44]: %timeit (df.reindex(columns=cols, index=idx))
1000 loops, best of 3: 402 µs per loop

仅索引：

In [49]: %timeit (df.reindex(idx))
The slowest run took 5.16 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 271 µs per loop

In [50]: %timeit (df.reindex_axis(idx))
The slowest run took 6.50 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 252 µs per loop


In [51]: %timeit (df.loc[['z', 'x', 'y']])
The slowest run took 5.51 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 418 µs per loop

In [52]: %timeit (df.loc[['z', 'x', 'y'], :])
The slowest run took 4.87 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 542 µs per loop

def pir(df):
    idx = ['z','x','y']
    a = df.index.values.searchsorted(idx)
    df = pd.DataFrame(
        df.values[a],
        df.index[a], df.columns
    )
    return df

In [63]: %timeit (pir(df))
The slowest run took 7.75 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 91.8 µs per loop

Answer 2

使用 (float)value 是一种非常自然的方法

loc

您可以使用

将其分配回数据框

df.loc[['z', 'x', 'y']]

A  d  a  b  c
B            
z  1  6  3  8
x  7  6  7  6
y  3  5  6  5

两个轴合在一起df = df.loc[['z', 'x', 'y']]

loc

使用 df.loc[['z', 'x', 'y'], ['d', 'a', 'b', 'c']] A d a b c B z 1 6 3 8 x 7 6 7 6 y 3 5 6 5

快速完成此操作

numpy.searchsorted

熊猫：有一种通过提供索引标签列表来排序行的本地方法吗？

2 个答案: