itj*_*s18 44 python sorting pandas
阅读完:http://pandas.pydata.org/pandas-docs/version/0.13.1/generated/pandas.DataFrame.sort.html
我似乎还无法弄清楚如何通过自定义列表对列进行排序.显然,默认排序是按字母顺序排列的.我举个例子.这是我的(非常删节)数据框:
Player Year Age Tm G
2967 Cedric Hunter 1991 27 CHH 6
5335 Maurice Baker 2004 25 VAN 7
13950 Ratko Varda 2001 22 TOT 60
6141 Ryan Bowen 2009 34 OKC 52
6169 Adrian Caldwell 1997 31 DAL 81
我希望能够按Player,Year和Tm进行排序.按正常顺序,播放器和年份的默认排序对我来说没问题.但是,我不希望团队按字母顺序排序b/c我希望TOT始终位于顶部.
这是我创建的列表:
sorter = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL', 'DEN',
'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN',
'WAS', 'WSB']
阅读完上面的链接后,我认为这样可行,但事实并非如此:
df.sort(['Player', 'Year', 'Tm'], ascending = [True, True, sorter])
它仍然在顶部有ATL,这意味着它按字母顺序排序,而不是根据我的自定义列表.任何帮助真的会非常感激,我只是想不出来.
dme*_*meu 59
我刚刚发现使用pandas 15.1可以使用分类系列(http://pandas.pydata.org/pandas-docs/stable/10min.html#categoricals)
至于您的示例,我们定义相同的数据框架和分拣机:
import pandas as pd
data = {
'id': [2967, 5335, 13950, 6141, 6169],
'Player': ['Cedric Hunter', 'Maurice Baker',
'Ratko Varda' ,'Ryan Bowen' ,'Adrian Caldwell'],
'Year': [1991, 2004, 2001, 2009, 1997],
'Age': [27, 25, 22, 34, 31],
'Tm': ['CHH', 'VAN', 'TOT', 'OKC', 'DAL'],
'G': [6, 7, 60, 52, 81]
}
# Create DataFrame
df = pd.DataFrame(data)
# Define the sorter
sorter = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL', 'DEN',
'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN', 'WAS', 'WSB']
使用数据框架和分拣机(类别订单),我们可以在pandas 15.1中执行以下操作:
# Convert Tm-column to category and in set the sorter as categories hierarchy
# Youc could also do both lines in one just appending the cat.set_categories()
df.Tm = df.Tm.astype("category")
df.Tm.cat.set_categories(sorter, inplace=True)
print(df.Tm)
Out[48]:
0 CHH
1 VAN
2 TOT
3 OKC
4 DAL
Name: Tm, dtype: category
Categories (38, object): [TOT < ATL < BOS < BRK ... UTA < VAN < WAS < WSB]
df.sort_values(["Tm"]) ## 'sort' changed to 'sort_values'
Out[49]:
Age G Player Tm Year id
2 22 60 Ratko Varda TOT 2001 13950
0 27 6 Cedric Hunter CHH 1991 2967
4 31 81 Adrian Caldwell DAL 1997 6169
3 34 52 Ryan Bowen OKC 2009 6141
1 25 7 Maurice Baker VAN 2004 5335
Mar*_*ius 31
从版本 1.1.0 开始，您可以使用该key属性对值进行排序：
df.sort_values(by="Tm", key=lambda column: column.map(lambda e: sorter.index(e)), inplace=True)
Gui*_*not 25
下面是一个对数据帧执行字典排序的示例.我们的想法是根据特定的排序创建一个数字索引.然后根据索引执行数字排序.将一列添加到数据框中,然后将其删除.
import pandas as pd
# Create DataFrame
df = pd.DataFrame(
{'id':[2967, 5335, 13950, 6141, 6169],\
'Player': ['Cedric Hunter', 'Maurice Baker' ,\
'Ratko Varda' ,'Ryan Bowen' ,'Adrian Caldwell'],\
'Year': [1991 ,2004 ,2001 ,2009 ,1997],\
'Age': [27 ,25 ,22 ,34 ,31],\
'Tm':['CHH' ,'VAN' ,'TOT' ,'OKC' ,'DAL'],\
'G':[6 ,7 ,60 ,52 ,81]})
# Define the sorter
sorter = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL','DEN',\
'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',\
'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',\
'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN',\
'WAS', 'WSB']
# Create the dictionary that defines the order for sorting
sorterIndex = dict(zip(sorter,range(len(sorter))))
# Generate a rank column that will be used to sort
# the dataframe numerically
df['Tm_Rank'] = df['Tm'].map(sorterIndex)
# Here is the result asked with the lexicographic sort
# Result may be hard to analyze, so a second sorting is
# proposed next
## NOTE:
## Newer versions of pandas use 'sort_values' instead of 'sort'
df.sort_values(['Player', 'Year', 'Tm_Rank'], \
ascending = [True, True, True], inplace = True)
df.drop('Tm_Rank', 1, inplace = True)
print(df)
# Here is an example where 'Tm' is sorted first, that will
# give the first row of the DataFrame df to contain TOT as 'Tm'
df['Tm_Rank'] = df['Tm'].map(sorterIndex)
## NOTE:
## Newer versions of pandas use 'sort_values' instead of 'sort'
df.sort_values(['Tm_Rank', 'Player', 'Year'], \
ascending = [True , True, True], inplace = True)
df.drop('Tm_Rank', 1, inplace = True)
print(df)
Arl*_*leg 11
根据pandas 1.1.0 文档，可以key像在sorted函数中一样使用参数进行排序（终于！）。在这里我们如何排序Tm
import pandas as pd
data = {
'id': [2967, 5335, 13950, 6141, 6169],
'Player': ['Cedric Hunter', 'Maurice Baker',
'Ratko Varda' ,'Ryan Bowen' ,'Adrian Caldwell'],
'Year': [1991, 2004, 2001, 2009, 1997],
'Age': [27, 25, 22, 34, 31],
'Tm': ['CHH', 'VAN', 'TOT', 'OKC', 'DAL'],
'G': [6, 7, 60, 52, 81]
}
# Create DataFrame
df = pd.DataFrame(data)
def tm_sorter(column):
"""Sort function"""
teams = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL', 'DEN',
'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN',
'WAS', 'WSB']
correspondence = {team: order for order, team in enumerate(teams)}
return column.map(correspondence)
df.sort_values(by='Tm', key=tm_sorter)
可悲的是，看起来我们只能在按 1 列排序时使用此功能（带keys 的列表是不可接受的）。可以通过以下方式规避groupby
df.sort_values(['Player', 'Year']) \
.groupby(['Player', 'Year']) \
.apply(lambda x: x.sort_values(by='Tm', key=tm_sorter)) \
.reset_index(drop=True)
如果您知道如何key在sort_values多列中使用，请告诉我
DataFrame.loc当您需要按单个自定义列表进行排序时，设置索引非常有用。因为loc将为不在 DataFrame 中的NaN值创建行，sorter我们将首先找到交集。这可以防止任何不需要的向上转换。任何值不在列表中的行都将被删除。
true_sort = [s for s in sorter if s in df.Tm.unique()]
df = df.set_index('Tm').loc[true_sort].reset_index()
Tm id Player Year Age G
0 TOT 13950 Ratko Varda 2001 22 60
1 CHH 2967 Cedric Hunter 1991 27 6
2 DAL 6169 Adrian Caldwell 1997 31 81
3 OKC 6141 Ryan Bowen 2009 34 52
4 VAN 5335 Maurice Baker 2004 25 7
起始数据：
print(df)
id Player Year Age Tm G
0 2967 Cedric Hunter 1991 27 CHH 6
1 5335 Maurice Baker 2004 25 VAN 7
2 13950 Ratko Varda 2001 22 TOT 60
3 6141 Ryan Bowen 2009 34 OKC 52
4 6169 Adrian Caldwell 1997 31 DAL 81
sorter = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL', 'DEN',
'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN', 'WAS', 'WSB']
小智 9
这只需几行即可完成工作
# Create a dummy df with the required list and the col name to sort on
dummy = pd.Series(sort_list, name = col_name).to_frame()
# Use left merge on the dummy to return a sorted df
sorted_df = pd.merge(dummy, df, on = col_name, how = 'left')