Python 使用pandas按列返回n个最小索引
我有以下(简化的)数据帧:Python 使用pandas按列返回n个最小索引,python,pandas,Python,Pandas,我有以下(简化的)数据帧: df = pd.DataFrame({'X': [1, 2, 3, 4, 5,6,7,8,9,10], 'Y': [10,20,30,40,50,-10,-20,-30,-40,-50], 'Z': [20,18,16,14,12,10,8,6,4,2]},index=list('ABCDEFGHIJ')) 其中给出了以下内容: X Y Z A 1 10 20 B 2 20 18 C 3 30 16 D 4 40 14
df = pd.DataFrame({'X': [1, 2, 3, 4, 5,6,7,8,9,10],
'Y': [10,20,30,40,50,-10,-20,-30,-40,-50],
'Z': [20,18,16,14,12,10,8,6,4,2]},index=list('ABCDEFGHIJ'))
其中给出了以下内容:
X Y Z
A 1 10 20
B 2 20 18
C 3 30 16
D 4 40 14
E 5 50 12
F 6 -10 10
G 7 -20 8
H 8 -30 6
I 9 -40 4
J 10 -50 2
我想创建一个新的dataframe,它按列返回n个最小值的索引
期望输出(例如,3个最小值):
执行此操作的最佳方法是什么?您可以将
应用于:
更快的numpy解决方案,包括:
计时:
In [111]: %timeit (pd.DataFrame(df.index[np.argsort(-df.values, axis=0)[-1:-1-N:-1]], columns=df.columns))
159 µs ± 1.37 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [112]: %timeit (df.apply(lambda x: pd.Series(x.nsmallest(N).index)))
3.52 ms ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
首先,您希望对每列的输入数据帧进行排序,然后获取每列的所有索引的列表,从这些索引创建一个数据帧,然后从结果数据帧返回前n行
def topN(df, n):
#first, sort dataframe per column
sort_x = df.sort_values(by = ['X'], ascending = True)
sort_y = df.sort_values(by = ['Y'], ascending = True)
sort_z = df.sort_values(by = ['Z'], ascending = True)
#now get a list of the indices of each sorted df
index_list_x = sort_x.index.values.tolist()
index_list_y = sort_y.index.values.tolist()
index_list_z = sort_z.index.values.tolist()
#create dataframe from lists
sorted_df = pd.DataFrame(
{'sorted_x':index_list_x,
'sorted_y':index_list_y,
'sorted_z':index_list_z
})
#return the top n from the sorted dataframe
return sorted_df.iloc[0:n]
topN(df,3)
返回:
X Y Z
0 A J J
1 B I I
2 C H H
In [111]: %timeit (pd.DataFrame(df.index[np.argsort(-df.values, axis=0)[-1:-1-N:-1]], columns=df.columns))
159 µs ± 1.37 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [112]: %timeit (df.apply(lambda x: pd.Series(x.nsmallest(N).index)))
3.52 ms ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
def topN(df, n):
#first, sort dataframe per column
sort_x = df.sort_values(by = ['X'], ascending = True)
sort_y = df.sort_values(by = ['Y'], ascending = True)
sort_z = df.sort_values(by = ['Z'], ascending = True)
#now get a list of the indices of each sorted df
index_list_x = sort_x.index.values.tolist()
index_list_y = sort_y.index.values.tolist()
index_list_z = sort_z.index.values.tolist()
#create dataframe from lists
sorted_df = pd.DataFrame(
{'sorted_x':index_list_x,
'sorted_y':index_list_y,
'sorted_z':index_list_z
})
#return the top n from the sorted dataframe
return sorted_df.iloc[0:n]
topN(df,3)
X Y Z
0 A J J
1 B I I
2 C H H