过滤序列中包含特定值的df-pandas

Question

我有一个对df进行子集设置的复杂过程。对于给定的序列或行，我希望返回特定的值。具体地，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位于给定序列内，我想为同一序列返回Up或Down。如果在包含X或Y的序列中未找到Up或Down，则返回Left或Right。如果未找到Up或Down、X或Y，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

这里有一种方法：

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，使用cumsum和startswith'Start_'创建组grp
• 接下来，创建三个布尔序列，分别标记为"X Y"、"上下"one_answers"左"对
• 然后，创建一个自定义函数，接受每个组，如果该组包含"X_Y"的True记录，然后构建数据帧按顺序串联"X_Y"、"Up_Down"one_answers"Left_Right"。使用head(2(只得到每组中的前两个记录
• 从生成结果数据帧后删除辅助列分组

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；希望我能清楚地解释你的挑战：

# 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