将最后一行的值添加到此行

Question

我想在按名称分组时获取最后一行的值。例如，在第2行中Walter这个名字的最后一次迭代中，我想得到Dog+“，”+猫代表Col1，啤酒+“，”+葡萄酒代表Col3。有很多列，所以我想让它基于索引/列位置，而不是列名

+------+---------+-------+
| Col1 |  Name   | Col3  |
+------+---------+-------+
| Dog  | Walter  | Beer  |
| Cat  | Walter  | Wine  |
| Dog  | Alfonso | Cider |
| Dog  | Alfonso | Cider |
| Dog  | Alfonso | Vodka |
+------+---------+-------+

这是我想要的输出：

+---------------+---------------------------+---------------------+
|     Col1      |           Name            |        Col3         |
+---------------+---------------------------+---------------------+
| Dog           | Walter                    | Beer                |
| Dog, Cat      | Walter, Walter            | Beer, Wine          |
| Dog           | Alfonso                   | Cider               |
| Dog, Dog      | Alfonso, Alfonso          | Cider, Cider        |
| Dog, Dog, Dog | Alfonso, Alfonso, Alfosno | Cider, Cider, Vodka |
+---------------+---------------------------+---------------------+

这是我尝试过的（但不起作用）：

for i in df:
    if df.loc[i,1] == df.loc[i+1,1]:
        df.loc[i,0] + ", " + df.loc[i+1,0]
    else:
        df.loc[i+1,0]

我读到不赞成使用for循环在pandas中的行上迭代，因此我希望通过使用矢量化或apply（或其他有效的方法）获得输出

Answer 1

下面是在索引上使用accumulate和使用df.agg方法的另一种方法：

from itertools import accumulate
import numpy as np

def fun(a):
    l = [[i] for i in a.index]
    acc = list(accumulate(l, lambda x, y: np.concatenate([x, y])))
    return pd.concat([a.loc[idx].agg(','.join) for idx in acc],axis=1).T
out = pd.concat([fun(v) for k,v in df.groupby('Name',sort=False)])

---

print(out)
          Col1                     Name               Col3
0          Dog                   Walter               Beer
1      Dog,Cat            Walter,Walter          Beer,Wine
0          Dog                  Alfonso              Cider
1      Dog,Dog          Alfonso,Alfonso        Cider,Cider
2  Dog,Dog,Dog  Alfonso,Alfonso,Alfonso  Cider,Cider,Vodka

您可以在最后添加一个带有drop=True的重置索引来重置索引

Answer 2

您可以使用groupby和cumsum。如果您不介意（取决于您在之后的使用）在末尾添加一个额外的逗号/空格，您可以执行以下操作：

print (df.groupby('Name')[['Col1', 'Col3']].apply(lambda x: (x + ', ').cumsum()))
              Col1                   Col3
0            Dog,                  Beer, 
1       Dog, Cat,            Beer, Wine, 
2            Dog,                 Cider, 
3       Dog, Dog,          Cider, Cider, 
4  Dog, Dog, Dog,   Cider, Cider, Vodka, 

但是，如果要删除额外的逗号/空格，只需在每列中添加str[：-2]，如下所示：

print (df.groupby('Name')[['Col1', 'Col3']].apply(lambda x: (x + ', ').cumsum())\
         .apply(lambda x: x.str[:-2]))
            Col1                 Col3
0            Dog                 Beer
1       Dog, Cat           Beer, Wine
2            Dog                Cider
3       Dog, Dog         Cider, Cider
4  Dog, Dog, Dog  Cider, Cider, Vodka

Answer 3

基本上，您要做的是在每个组上运行一个交换聚合函数。Pandas有comsum用于常规加法，但不支持自定义交换函数。为此，您可能需要使用一些numpy函数：

df = pd.DataFrame({"col1": ["D", "C", "D", "D", "D"], "Name": ["W", "W", "A", "A", "A"], 
                   "col3": ["B", "W", "C", "C", "V"] })


import numpy as np
def ser_accum(op,ser):
    u_op = np.frompyfunc(op, 2, 1) # two inputs, one output
    return u_op.accumulate(ser, dtype=np.object)

def plus(x,y):
    return x + "," + y

def accum(df):
    for col in df.columns:
        df[col] = ser_accum(plus, df[col])
    return df

df.groupby("Name").apply(accum)

结果如下：

col1    Name    col3
0   D   W   B
1   D,C W,W B,W
2   D   A   C
3   D,D A,A C,C
4   D,D,D   A,A,A   C,C,V