Question

我有一个有点复杂的子集 df 过程。对于给定的序列或行，我想返回特定的值。具体来说，使用下面的 df，序列的开始由 Start_A、Start_B、Start_C 表示。如果 X 或 Y 位于给定序列中，我想为同一序列返回 Up 或 Down。如果在包含 Up 或 Down 的序列中未找到 X 或 Y，则返回 Left 或 Right。如果 Up 或 Down 或 X 或 Y 均未找到，则打印错误。

import pandas as pd

df = pd.DataFrame({      
    'Num' : [1,2,3,4,6,7,9,10,12,13,14,15,17,18,19,21,22,23,25,26,27,28,30,31,32],
    'Item' : ['Start_A','AB','CD','Left','Start_C','CD','X','Up','Right','Start_C','EF','AB','Y','AB','Down','Left','Start_B','AB','Y','CD','Left','Start_A','AB','CD','Right'],        
    })
  
m1 = df['Item'].isin(['X','Y']).cumsum().gt(0)
m2 = df['Item'].isin(['Up','Down']).iloc[::-1].cumsum().gt(0)

df1 = df[m1 & m2]

原始 df：

    Num     Item
0     1  Start_A # No X,Y within sequence. drop all
1     2       AB
2     3       CD
3     4     Left
4     6  Start_C # X and Up within sequence.
5     7       CD
6     9        X
7    10       Up
8    12    Right
9    13  Start_C # Y and Down within sequence.
10   14       EF
11   15       AB
12   17        Y
13   18       AB
14   19     Down
15   21     Left
16   22  Start_B # Y within sequence. No Up/Down. But Left is.
17   23       AB
18   25        Y
19   26       CD
20   27       AB
21   27     Left
22   28  Start_A # No X,Y within sequence. drop all
23   30       AB
24   31       CD
25   32    Right

预期输出：

    Num     Item
6     9        X
7    10       Up
12   17        Y
14   19     Down
18   25        Y
21   27     Left

Answer 1

这是一种方法：

import pandas as pd
import numpy as np

df = pd.DataFrame({      
    'Num' : [1,2,3,4,6,7,9,10,12,13,14,15,17,18,19,21,22,23,25,26,27,28,30,31,32],
    'Item' : ['Start_A','AB','CD','Left','Start_C','CD','X','Up','Right','Start_C','EF','AB','Y','AB','Down','Left','Start_B','AB','Y','CD','Left','Start_A','AB','CD','Right'],        
    })
  

grp = df['Item'].str.startswith('Start_').cumsum()

df['X_Y'] = df['Item'].isin(['X', 'Y'])
df['Up_Down'] = df['Item'].isin(['Up', 'Down'])
df['Left_Right'] = df['Item'].isin(['Left', 'right'])

def f(x):
    if x['X_Y'].any():
        return pd.concat([x[x['X_Y']], x[x['Up_Down']], x[x['Left_Right']]]).head(2)

df.groupby(grp, group_keys=False).apply(f).drop(['X_Y', 'Up_Down', 'Left_Right'], axis=1)

输出：

    Num  Item
6     9     X
7    10    Up
12   17     Y
14   19  Down
18   25     Y
20   27  Left

详细信息：

首先，创建组 grp，使用 cumsum 并以“Start_”开头。
接下来，创建三个布尔系列标记为“X Y”、“Up Down”和“Left” 对'
然后，创建一个接受每个组的自定义函数，如果该组包含“X_Y”的真实记录，然后构建一个数据框按顺序连接“X_Y”、“Up_Down”和“Left_Right”。采用 head(2) 只获取每组的前两条记录。
从构建结果数据框后删除辅助列

Answer 2

m=df['Item'].str.contains('Start').cumsum()#Create mask
g=df[df.groupby(m)['Item'].transform(lambda x:x.str.contains('X|Y').any())]#new dataframe where each occurance must have X or Y
t=g['Item'].where(g['Item'].str.contains('X|Y|Left|Right',np.nan))#replace any values not included
s=g['Item'].where(g['Item'].str.contains('X|Y|Up|Down',np.nan))#replace any values not included
g=g.assign(drop=np.where(g.groupby(m)['Item'].transform(lambda x:x.isin(['Up','Down']).any()),s,t)).dropna()#Use np.where to enforce contions s and t
print(g)



 Num  Item  drop
6     9     X     X
7    10    Up    Up
12   17     Y     Y
14   19  Down  Down
18   25     Y     Y
20   27  Left  Left

Answer 3

在代码中添加了注释；希望你能清楚我是如何推理你的挑战的：

 # thanks to Scott Boston for a simpler syntax here
(df.assign(counter = df.Item.str.startswith("Start_").cumsum(), 
           boolean = lambda df: df.groupby('counter').transform(",".join), 
           #first phase, X or Y should be present
           # if absent, nulls will be introduced
           boolean_1 = lambda df: df.boolean.str.extract(r"(X|Y)")
           )
   .dropna()
    # next phase, get them in order of Up, Down, Left, Right
    # use extract, since it returns the first match
   .assign(boolean_2 = lambda df: df.boolean
                                    .str.extract(r"(Up|Down|Left|Right)"))
    # filter and keep the original columns
   .query("Item == boolean_1 or Item == boolean_2")
   .filter(['Num', 'Item'])
)


    Num  Item
6     9     X
7    10    Up
12   17     Y
14   19  Down
18   25     Y
20   27  Left

过滤包含序列中特定值的 df - pandas

3 个答案: