用排序索引替換 pandas 列

Question

我有一個示例 DF，試圖用升序索引替換列值列表：

東風：

df = pd.DataFrame(np.random.randint(0,10,size=(7,3)),columns=["a","b","c"])
df["d1"]=["Apple","Mango","Apple","Mango","Mango","Mango","Apple"]
df["d2"]=["Orange","lemon","lemon","Orange","lemon","Orange","lemon"]
df["date"] = ["2002-01-01","2002-01-01","2002-01-01","2002-01-01","2002-02-01","2002-02-01","2002-02-01"]
df["date"] = pd.to_datetime(df["date"])

    a   b   c    d1      d2       date
0   2   7   9   Apple   Orange  2002-01-01
1   6   0   9   Mango   lemon   2002-01-01
2   8   0   0   Apple   lemon   2002-01-01
3   4   4   4   Mango   Orange  2002-01-01
4   5   0   8   Mango   lemon   2002-02-01
5   6   1   6   Mango   Orange  2002-02-01
6   7   2   7   Apple   lemon   2002-02-01

步驟1：

Group the DF by "date" column, sample group on "2002-01-01"


        a   b   c    d1      d2       date
    0   2   7   9   Apple   Orange  2002-01-01
    1   6   0   9   Mango   lemon   2002-01-01
    2   8   0   0   Apple   lemon   2002-01-01
    3   4   4   4   Mango   Orange  2002-01-01

第2步：

在該組中，將列["d1","d2"]的值替換為基於c的排序平均值的索引（不是 DF 索引）。

例如在上面的組中mean(c, d1="Apple") = [9+0]/2 => 4.5和mean(c, d1="Mango") = [9+4]/2 => 6.5所以ascending sorted index是Apple:0和Mango:1

所以列d1的值將被替換如下：

            a   b   c   d1       d2       date
        0   2   7   9   0      Orange   2002-01-01
        1   6   0   9   1      lemon    2002-01-01
        2   8   0   0   0      lemon    2002-01-01
        3   4   4   4   1      Orange   2002-01-01

將此應用於整個df 。 我有一種遍歷組和每一行的蠻力方法，任何關於更多基於pandas的解決方案的建議都將有助於提高效率。

Answer 1

這是您在 d1 列中尋找的內容嗎？ 您也可以對 d2 應用一些類似的技術。 雖然它不是最優雅的解決方案。

import pandas as pd
import numpy as np

df = pd.DataFrame(np.random.randint(0,10,size=(7,3)),columns=["a","b","c"])
df["d1"]=["Apple","Mango","Apple","Mango","Mango","Mango","Apple"]
df["d2"]=["Orange","lemon","lemon","Orange","lemon","Orange","lemon"]
df["date"] = ["2002-01-01","2002-01-01","2002-01-01","2002-01-01","2002-02-01","2002-02-01","2002-02-01"]
df["date"] = pd.to_datetime(df["date"])

df['mean_value'] = df.groupby(['date', 'd1'])['c'].transform(lambda x: np.mean(x))
df['rank_value'] = (df.groupby(['date'])['mean_value'].rank(ascending=True, method='dense') - 1).astype(int)
df['d1'] = df['rank_value']
df.drop(labels=['rank_value', 'mean_value'], axis=1, inplace=True)

df

   a  b  c  d1      d2       date
0  3  1  4   1  Orange 2002-01-01
1  9  7  5   0   lemon 2002-01-01
2  9  9  5   1   lemon 2002-01-01
3  8  1  2   0  Orange 2002-01-01
4  8  0  1   0   lemon 2002-02-01
5  1  8  3   0  Orange 2002-02-01
6  8  0  4   1   lemon 2002-02-01

Answer 2

您可以使用pivot_table和groupby.rank創建排名。 之后使用map重新分配值

df1 = df.pivot_table('c', ['date','d1']).groupby(level=0).rank(method='dense')-1
df['d1'] = df[['date','d1']].agg(tuple, axis=1).map(df1.c).astype('int')

Out[255]:
   a  b  c  d1      d2        date
0  2  7  9   0  Orange  2002-01-01
1  6  0  9   1   lemon  2002-01-01
2  8  0  0   0   lemon  2002-01-01
3  4  4  4   1  Orange  2002-01-01
4  5  0  8   0   lemon  2002-02-01
5  6  1  6   0  Orange  2002-02-01
6  7  2  7   0   lemon  2002-02-01

注意：組2002-02-01對於Mango和Apple具有相同的平均值7 ，因此排名均為0