我有一个数据框，像这样：
print(df)
0 1 2
0 354.7 April 4.0
1 55.4 August 8.0
2 176.5 December 12.0
3 95.5 February 2.0
4 85.6 January 1.0
5 152 July 7.0
6 238.7 June 6.0
7 104.8 March 3.0
8 283.5 May 5.0
9 278.8 November 11.0
10 249.6 October 10.0
11 212.7 September 9.0
如您所见，月份并没有按照日历的顺序排列。因此我创建了第二列来获取每个月对应的月份数字（1-12）。从那里开始，我该如何根据日历月份的顺序对这个数据框进行排序？
使用sort_values按特定列的值对数据框进行排序：
In [18]:
df.sort_values('2')
Out[18]:
0 1 2
4 85.6 January 1.0
3 95.5 February 2.0
7 104.8 March 3.0
0 354.7 April 4.0
8 283.5 May 5.0
6 238.7 June 6.0
5 152.0 July 7.0
1 55.4 August 8.0
11 212.7 September 9.0
10 249.6 October 10.0
9 278.8 November 11.0
2 176.5 December 12.0
sort_values，其中列标签根据排序优先级进行排序。如果使用df.sort_values(['2', '0'])，则结果将按列2然后按列0排序。尽管如此，在这个例子中这样做并没有什么实际意义，因为df['2']中的每个值都是唯一的。ascending=False是将数据框按降序排序，默认情况下为True。我使用的是Python 3.6.6和Pandas 0.23.4版本。final_df = df.sort_values(by=['2'], ascending=False)
您可以在pandas文档这里中查看更多细节。
使用列名对我有效。
sorted_df = df.sort_values(by=['Column_name'], ascending=True)
ascending（bool或bool列表）：
由于默认为升序，且OP的目标是升序排序，因此不需要指定该参数（有关解决降序的方法，请参见下面的最后一条注释），因此可以使用以下其中一种方式：升序还是降序。指定多个排序顺序的列表。如果这是布尔列表，则必须与by的长度匹配。
Performing the operation in-place, and keeping the same variable name. This requires one to pass inplace=True as follows:
df.sort_values(by=['2'], inplace=True)
# or
df.sort_values(by = '2', inplace = True)
# or
df.sort_values('2', inplace = True)
If doing the operation in-place is not a requirement, one can assign the change (sort) to a variable:
With the same name of the original dataframe, df as
df = df.sort_values(by=['2'])
With a different name, such as df_new, as
df_new = df.sort_values(by=['2'])
0 1 2
4 85.6 January 1.0
3 95.5 February 2.0
7 104.8 March 3.0
0 354.7 April 4.0
8 283.5 May 5.0
6 238.7 June 6.0
5 152 July 7.0
1 55.4 August 8.0
11 212.7 September 9.0
10 249.6 October 10.0
9 278.8 November 11.0
2 176.5 December 12.0
最后，可以使用pandas.DataFrame.reset_index重置索引，得到以下结果：
df.reset_index(drop = True, inplace = True)
# or
df = df.reset_index(drop = True)
[Out]:
0 1 2
0 85.6 January 1.0
1 95.5 February 2.0
2 104.8 March 3.0
3 354.7 April 4.0
4 283.5 May 5.0
5 238.7 June 6.0
6 152 July 7.0
7 55.4 August 8.0
8 212.7 September 9.0
9 249.6 October 10.0
10 278.8 November 11.0
11 176.5 December 12.0
df = df.sort_values(by=['2']).reset_index(drop = True)
[Out]:
0 1 2
0 85.6 January 1.0
1 95.5 February 2.0
2 104.8 March 3.0
3 354.7 April 4.0
4 283.5 May 5.0
5 238.7 June 6.0
6 152 July 7.0
7 55.4 August 8.0
8 212.7 September 9.0
9 249.6 October 10.0
10 278.8 November 11.0
11 176.5 December 12.0
注意：
If one is not doing the operation in-place, forgetting the steps mentioned above may lead one (as this user) to not be able to get the expected result.
There are strong opinions on using inplace. For that, one might want to read this.
One is assuming that the column 2 is not a string. If it is, one will have to convert it:
Using pandas.to_numeric
df['2'] = pd.to_numeric(df['2'])
Using pandas.Series.astype
df['2'] = df['2'].astype(float)
If one wants in descending order, one needs to pass ascending=False as
df = df.sort_values(by=['2'], ascending=False)
# or
df.sort_values(by = '2', ascending=False, inplace=True)
[Out]:
0 1 2
2 176.5 December 12.0
9 278.8 November 11.0
10 249.6 October 10.0
11 212.7 September 9.0
1 55.4 August 8.0
5 152 July 7.0
6 238.7 June 6.0
8 283.5 May 5.0
0 354.7 April 4.0
7 104.8 March 3.0
3 95.5 February 2.0
4 85.6 January 1.0
就像另一种解决方案：
不需要创建第二列，您可以对字符串数据（月份名称）进行分类，并像这样进行排序：
df.rename(columns={1:'month'},inplace=True)
df['month'] = pd.Categorical(df['month'],categories=['December','November','October','September','August','July','June','May','April','March','February','January'],ordered=True)
df = df.sort_values('month',ascending=False)
在创建 Categorical 对象时，它将按照你指定的月份名称为你提供有序的数据。
只需对数据进行一些额外的操作。假设我们有一个数据框df，我们可以执行几个操作以获得所需的输出。
ID cost tax label
1 216590 1600 test
2 523213 1800 test
3 250 1500 experiment
(df['label'].value_counts().to_frame().reset_index()).sort_values('label', ascending=False)
将会给出一个标签按照顺序排列的数据框输出
index label
0 test 2
1 experiment 1
这对我有效
df.sort_values(by='Column_name', inplace=True, ascending=False)
df = df.sort_values('2')
df = df.reset_index(drop=True)
df.sort_values('2', inplace=True)会完成这个任务。 - undefined以下是 Pandas 文档中关于 sort_values 的模板。
DataFrame.sort_values(by, axis=0,
ascending=True,
inplace=False,
kind='quicksort',
na_position='last',
ignore_index=False, key=None)[source]
在这种情况下，代码如下所示：
df.sort_values(by=['2'])
API 参考文献：pandas.DataFrame.sort_values
再添加一些见解
df=raw_df['2'].sort_values() # will sort only one column (i.e 2)
但是，
df =raw_df.sort_values(by=["2"] , ascending = False) # this will sort the whole df in decending order on the basis of the column "2"
ascending=False来进行降序排序 - undefined