如何根据其他两个列中的条件创建和填充新列？

Question

如何根据另外两个列的条件创建一个新列并用值填充值？

输入：

    import pandas as pd
    import numpy as np

    list1 = ['no','no','yes','yes','no','no','no','yes','no','yes','yes','no','no','no']
    list2 = ['no','no','no','no','no','yes','yes','no','no','no','no','no','yes','no']

    df = pd.DataFrame({'A' : list1, 'B' : list2}, columns = ['A', 'B'])

    df['C'] = np.where ((df['A'] == 'yes') & (df['A'].shift(1) == 'no'), 'X', np.nan)
    df['D'] = 'nan','nan','X','X','X','X','nan','X','X','X','X','X','X','nan'

    print (df)

输出：

          A    B    C    D
    0    no   no  nan  nan
    1    no   no  nan  nan
    2   yes   no    X    X
    3   yes   no  nan    X
    4    no   no  nan    X
    5    no  yes  nan    X
    6    no  yes  nan  nan
    7   yes   no    X    X
    8    no   no  nan    X
    9   yes   no    X    X
    10  yes   no  nan    X
    11   no   no  nan    X
    12   no  yes  nan    X
    13   no   no  nan  nan

A和B列将是Givens，仅包含“是”或“否”值。只能有三对（'no' - 'no'，'yes' - 'no'或'no' - 'YES'）。永远不可能有“是” - “是”对。

目的是在遇到“是” - “否”对时，将“ X”放在新列中，然后继续填充'X，直到有“否” - '是'是'对。这可能会在几行或几百行中发生。

D列显示了所需的输出。

C列是当前失败的尝试。

谁能帮忙？提前致谢。

Answer 1

尝试以下操作：

df["E"] = np.nan

# Use boolean indexing to set no-yes to placeholder value
df.loc[(df["A"] == "no") & (df["B"] == "yes"), "E"] = "PL"

# Shift placeholder down by one, as it seems from your example
# that you want X to be on the no-yes "stopping" row
df["E"] = df.E.shift(1)

# Then set the X value on the yes-no rows
df.loc[(df.A == "yes") & (df.B == "no"), "E"] = "X"
df["E"] = df.E.ffill() # Fill forward

# Fix placeholders
df.loc[df.E == "PL", "E"] = np.nan

结果：

    A   B   C   D   E
0   no  no  nan nan NaN
1   no  no  nan nan NaN
2   yes no  X   X   X
3   yes no  nan X   X
4   no  no  nan X   X
5   no  yes nan X   X
6   no  yes nan nan NaN
7   yes no  X   X   X
8   no  no  nan X   X
9   yes no  X   X   X
10  yes no  nan X   X
11  no  no  nan X   X
12  no  yes nan X   X
13  no  no  nan nan NaN

Answer 2

您可以使用Apply（）来做到这一点，

df['C'] = df[['A','B']].apply(yourfunction, axis=1)

您的功能可能是：

def yourfunction(cols):
   col_A = cols[0]
   col_B = cols[1]
   if YOURLOGIC:
      return X

Answer 3

您可以这样尝试。在这里，我使用iTerrows在行上循环

import pandas as pd
import numpy as np

list1 = ['no','no','yes','yes','no','no','no','yes','no','yes','yes','no','no','no']
list2 = ['no','no','no','no','no','yes','yes','no','no','no','no','no','yes','no']

df = pd.DataFrame({'A' : list1, 'B' : list2}, columns = ['A', 'B'])

df['C'] = np.nan
to_check = 0
for ind, row in df.iterrows():
    if (row['A'] == 'yes') and (row['B'] == 'no'):
        to_check = 1
        df.loc[ind, 'C'] = 'X'
        continue
    
    if (row['A'] == 'no') and (row['B'] == 'yes'):
        if to_check == 1:
            df.loc[ind, 'C'] = 'X'
            to_check = 0
        continue

    if to_check == 1:
        df.loc[ind, 'C'] = 'X'


df['D'] = 'nan','nan','X','X','X','X','nan','X','X','X','X','X','X','nan'

print (df)

Answer 4

这将完成工作，

def needed_in():
  count = False
  for index in df.index:
    if df.loc[index, ["A", "B"]].tolist() == ["yes", "no"]:
      count = True
    
    if count:
      yield index

    if df.loc[index, ["A", "B"]].tolist() == ["no", "yes"]:
      count = False

df["C"] = np.nan
df.loc[needed_in(), "C"] = "X"

输出 -

	a	b	c
0	否	否	nan
1	否	否	nan
2	是否	​	x
3	是否	​	​
​	​	​	​
​	​	​	​
​	​	​	​
​	​	​	​
​	​	​	​
​	​	​	​
​	​	​	​
​	​	​	x
12	否	是	x
13	否	否	nan