Python 熊猫数据框中的自定义排序
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Custom sorting in pandas dataframe
提问by Kathirmani Sukumar
I have python pandas dataframe, in which a column contains month name.
我有 python pandas 数据框，其中一列包含月份名称。
How can I do a custom sort using a dictionary, for example:
如何使用字典进行自定义排序，例如：
custom_dict = {'March':0, 'April':1, 'Dec':3}
采纳答案by Andy Hayden
Pandas 0.15 introduced Categorical Series, which allows a much clearer way to do this:
Pandas 0.15 引入了Categorical Series，它提供了一种更清晰的方法来做到这一点：
First make the month column a categorical and specify the ordering to use.
首先将月份列设为分类列并指定要使用的排序。
In [21]: df['m'] = pd.Categorical(df['m'], ["March", "April", "Dec"])
In [22]: df # looks the same!
Out[22]:
a b m
0 1 2 March
1 5 6 Dec
2 3 4 April
Now, when you sort the month column it will sort with respect to that list:
现在，当您对月份列进行排序时，它将根据该列表进行排序：
In [23]: df.sort_values("m")
Out[23]:
a b m
0 1 2 March
2 3 4 April
1 5 6 Dec
Note: if a value is not in the list it will be converted to NaN.
注意：如果一个值不在列表中，它将被转换为 NaN。
An older answer for those interested...
对于那些有兴趣的人来说，这是一个较旧的答案......
You could create an intermediary series, and set_indexon that:
您可以创建一个中间系列，然后set_index：
df = pd.DataFrame([[1, 2, 'March'],[5, 6, 'Dec'],[3, 4, 'April']], columns=['a','b','m'])
s = df['m'].apply(lambda x: {'March':0, 'April':1, 'Dec':3}[x])
s.sort_values()
In [4]: df.set_index(s.index).sort()
Out[4]:
a b m
0 1 2 March
1 3 4 April
2 5 6 Dec
As commented, in newer pandas, Series has a replacemethod to do this more elegantly:
正如评论的那样，在较新的熊猫中，Series 有一种replace方法可以更优雅地做到这一点：
s = df['m'].replace({'March':0, 'April':1, 'Dec':3})
The slight difference is that this won't raise if there is a value outside of the dictionary (it'll just stay the same).
略有不同的是，如果字典之外有值，则不会提高（它会保持不变）。
回答by eumiro
import pandas as pd
custom_dict = {'March':0,'April':1,'Dec':3}
df = pd.DataFrame(...) # with columns April, March, Dec (probably alphabetically)
df = pd.DataFrame(df, columns=sorted(custom_dict, key=custom_dict.get))
returns a DataFrame with columns March, April, Dec
返回一个包含三月、四月、十二月的 DataFrame
回答by Michael Delgado
Update: The accepted answeris now the right way to do this. Leaving my old answer below for posterity, but if you encounter this post - look above and propser.
更新：接受的答案现在是正确的方法。将我的旧答案留给后人，但如果您遇到这篇文章-请看上面和道具。
Original post
原帖
A bit late to the game, but here's a way to create a function that sorts pandas Series, DataFrame, and multiindex DataFrame objects using arbitrary functions.
游戏有点晚了，但这里有一种方法可以创建一个使用任意函数对 Pandas Series、DataFrame 和多索引 DataFrame 对象进行排序的函数。
I make use of the df.iloc[index]method, which references a row in a Series/DataFrame by position (compared to df.loc, which references by value). Using this, we just have to have a function that returns a series of positional arguments:
我使用该df.iloc[index]方法，该方法按位置引用 Series/DataFrame 中的一行（与df.loc按值引用的相比）。使用它，我们只需要一个返回一系列位置参数的函数：
def sort_pd(key=None,reverse=False,cmp=None):
def sorter(series):
series_list = list(series)
return [series_list.index(i)
for i in sorted(series_list,key=key,reverse=reverse,cmp=cmp)]
return sorter
You can use this to create custom sorting functions. This works on the dataframe used in Andy Hayden's answer:
您可以使用它来创建自定义排序功能。这适用于安迪海登的回答中使用的数据框：
df = pd.DataFrame([
[1, 2, 'March'],
[5, 6, 'Dec'],
[3, 4, 'April']],
columns=['a','b','m'])
custom_dict = {'March':0, 'April':1, 'Dec':3}
sort_by_custom_dict = sort_pd(key=custom_dict.get)
In [6]: df.iloc[sort_by_custom_dict(df['m'])]
Out[6]:
a b m
0 1 2 March
2 3 4 April
1 5 6 Dec
This also works on multiindex DataFrames and Series objects:
这也适用于多索引 DataFrames 和 Series 对象：
months = ['Jan','Feb','Mar','Apr','May','Jun','Jul','Aug','Sep','Oct','Nov','Dec']
df = pd.DataFrame([
['New York','Mar',12714],
['New York','Apr',89238],
['Atlanta','Jan',8161],
['Atlanta','Sep',5885],
],columns=['location','month','sales']).set_index(['location','month'])
sort_by_month = sort_pd(key=months.index)
In [10]: df.iloc[sort_by_month(df.index.get_level_values('month'))]
Out[10]:
sales
location month
Atlanta Jan 8161
New York Mar 12714
Apr 89238
Atlanta Sep 5885
sort_by_last_digit = sort_pd(key=lambda x: x%10)
In [12]: pd.Series(list(df['sales'])).iloc[sort_by_last_digit(df['sales'])]
Out[12]:
2 8161
0 12714
3 5885
1 89238
To me this feels clean, but it uses python operations heavily rather than relying on optimized pandas operations. I haven't done any stress testing but I'd imagine this could get slow on very large DataFrames. Not sure how the performance compares to adding, sorting, then deleting a column. Any tips on speeding up the code would be appreciated!
对我来说这感觉很干净，但它大量使用 python 操作而不是依赖优化的熊猫操作。我没有做过任何压力测试，但我想这在非常大的 DataFrame 上可能会变慢。不确定与添加、排序和删除列的性能相比如何。任何有关加速代码的提示将不胜感激！
回答by cs95
pandas >= 1.1
熊猫 >= 1.1
You will soon be able to use sort_valueswith keyargument:
您很快就可以使用sort_valueswithkey参数：
custom_dict = {'March': 0, 'April': 1, 'Dec': 3}
df
a b m
0 1 2 March
1 5 6 Dec
2 3 4 April
df.sort_values(key=lambda x: x.map(custom_dict))
a b m
0 1 2 March
2 3 4 April
1 5 6 Decpandas <= 1.0.X
熊猫 <= 1.0.X
One simple method is using the output Series.mapand Series.argsortto index into dfusing DataFrame.iloc(since argsort produces sorted integer positions); since you have a dictionary; this becomes easy.
一种简单的方法是使用输出Series.map并Series.argsort索引df使用DataFrame.iloc（因为 argsort 产生排序的整数位置）；因为你有字典；这变得容易。
df.iloc[df['m'].map(custom_dict).argsort()]
a b m
0 1 2 March
2 3 4 April
1 5 6 Dec
If you need to sort in descending order, invert the mapping.
如果您需要按降序排序，请反转映射。
df.iloc[(-df['m'].map(custom_dict)).argsort()]
a b m
1 5 6 Dec
2 3 4 April
0 1 2 March
Note that this only works on numeric items. Otherwise, you will need to workaround this using sort_values, and accessing the index:
请注意，这仅适用于数字项目。否则，您将需要使用sort_values, 并访问索引来解决此问题：
df.loc[df['m'].map(custom_dict).sort_values(ascending=False).index]
a b m
1 5 6 Dec
2 3 4 April
0 1 2 March
More options are available with astype(this is deprecated now), or pd.Categorical, but you need to specify ordered=Truefor it to work correctly.
更多的选择与astype（这是现在不建议使用），或者pd.Categorical，但你需要指定ordered=True为它工作正常。
# Older version,
# df['m'].astype('category',
# categories=sorted(custom_dict, key=custom_dict.get),
# ordered=True)
df['m'] = pd.Categorical(df['m'],
categories=sorted(custom_dict, key=custom_dict.get),
ordered=True)
Now, a simple sort_valuescall will do the trick:
现在，一个简单的sort_values调用就可以解决问题：
df.sort_values('m')
a b m
0 1 2 March
2 3 4 April
1 5 6 Dec
The categorical ordering will also be honoured when groupbysorts the output.
在groupby对输出进行排序时，也将遵循分类排序。
