问题:’DataFrame’对象没有属性’sort’

我在这里遇到一些问题,在我的python包中,我已经安装了numpy,但是我仍然遇到此错误‘DataFrame’对象没有属性’sort’

任何人都可以给我一些想法。

这是我的代码:

final.loc[-1] =['', 'P','Actual']
final.index = final.index + 1  # shifting index
final = final.sort()
final.columns=[final.columns,final.iloc[0]]
final = final.iloc[1:].reset_index(drop=True)
final.columns.names = (None, None)

I face some problem here, in my python package I have install numpy, but I still have this error ‘DataFrame’ object has no attribute ‘sort’

Anyone can give me some idea..

This is my code :

final.loc[-1] =['', 'P','Actual']
final.index = final.index + 1  # shifting index
final = final.sort()
final.columns=[final.columns,final.iloc[0]]
final = final.iloc[1:].reset_index(drop=True)
final.columns.names = (None, None)

回答 0

sort() 不推荐使用DataFrames,而采用以下任何一种方法:

sort()在Pandas中已弃用(但仍可用)版本0.17(2015-10-09),并引入sort_values()sort_index()。它已从0.20版(2017-05-05)的Pandas中删除。

sort() was deprecated for DataFrames in favor of either:

sort() was deprecated (but still available) in Pandas with release 0.17 (2015-10-09) with the introduction of sort_values() and sort_index(). It was removed from Pandas with release 0.20 (2017-05-05).


回答 1

熊猫排序101

sort已经在v0.20替换DataFrame.sort_valuesDataFrame.sort_index。除此之外,我们还有argsort

以下是一些常见的排序用例,以及如何使用当前API中的排序功能解决它们。首先,设置。

# Setup
np.random.seed(0)
df = pd.DataFrame({'A': list('accab'), 'B': np.random.choice(10, 5)})    
df                                                                                                                                        
   A  B
0  a  7
1  c  9
2  c  3
3  a  5
4  b  2

按单列排序

例如,要按df列“ A” 排序,请使用sort_values单个列名:

df.sort_values(by='A')

   A  B
0  a  7
3  a  5
4  b  2
1  c  9
2  c  3

如果您需要新的RangeIndex,请使用DataFrame.reset_index

按多列排序

例如,通过排序两个关口“A”和“B”中df,你可以通过一个列表sort_values

df.sort_values(by=['A', 'B'])

   A  B
3  a  5
0  a  7
4  b  2
2  c  3
1  c  9

按DataFrame索引排序

df2 = df.sample(frac=1)
df2

   A  B
1  c  9
0  a  7
2  c  3
3  a  5
4  b  2

您可以使用sort_index

df2.sort_index()

   A  B
0  a  7
1  c  9
2  c  3
3  a  5
4  b  2

df.equals(df2)                                                                                                                            
# False
df.equals(df2.sort_index())                                                                                                               
# True

以下是一些可比较的方法及其性能:

%timeit df2.sort_index()                                                                                                                  
%timeit df2.iloc[df2.index.argsort()]                                                                                                     
%timeit df2.reindex(np.sort(df2.index))                                                                                                   

605 µs ± 13.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
610 µs ± 24.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
581 µs ± 7.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

按指数列表排序

例如,

idx = df2.index.argsort()
idx
# array([0, 7, 2, 3, 9, 4, 5, 6, 8, 1])

这个“排序”问题实际上是一个简单的索引问题。仅传递整数标签即可iloc

df.iloc[idx]

   A  B
1  c  9
0  a  7
2  c  3
3  a  5
4  b  2

Pandas Sorting 101

sort has been replaced in v0.20 by DataFrame.sort_values and DataFrame.sort_index. Aside from this, we also have argsort.

Here are some common use cases in sorting, and how to solve them using the sorting functions in the current API. First, the setup.

# Setup
np.random.seed(0)
df = pd.DataFrame({'A': list('accab'), 'B': np.random.choice(10, 5)})    
df                                                                                                                                        
   A  B
0  a  7
1  c  9
2  c  3
3  a  5
4  b  2

Sort by Single Column

For example, to sort df by column “A”, use sort_values with a single column name:

df.sort_values(by='A')

   A  B
0  a  7
3  a  5
4  b  2
1  c  9
2  c  3

If you need a fresh RangeIndex, use DataFrame.reset_index.

Sort by Multiple Columns

For example, to sort by both col “A” and “B” in df, you can pass a list to sort_values:

df.sort_values(by=['A', 'B'])

   A  B
3  a  5
0  a  7
4  b  2
2  c  3
1  c  9

Sort By DataFrame Index

df2 = df.sample(frac=1)
df2

   A  B
1  c  9
0  a  7
2  c  3
3  a  5
4  b  2

You can do this using sort_index:

df2.sort_index()

   A  B
0  a  7
1  c  9
2  c  3
3  a  5
4  b  2

df.equals(df2)                                                                                                                            
# False
df.equals(df2.sort_index())                                                                                                               
# True

Here are some comparable methods with their performance:

%timeit df2.sort_index()                                                                                                                  
%timeit df2.iloc[df2.index.argsort()]                                                                                                     
%timeit df2.reindex(np.sort(df2.index))                                                                                                   

605 µs ± 13.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
610 µs ± 24.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
581 µs ± 7.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Sort by List of Indices

For example,

idx = df2.index.argsort()
idx
# array([0, 7, 2, 3, 9, 4, 5, 6, 8, 1])

This “sorting” problem is actually a simple indexing problem. Just passing integer labels to iloc will do.

df.iloc[idx]

   A  B
1  c  9
0  a  7
2  c  3
3  a  5
4  b  2

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