python - 根据其他列的值将新列添加到数据框

Question

我读过很多有关此问题的问题，但我不知道如何将这些解决方案应用于我自己的问题。我想我必须使用 df.apply 或 df.assign 但即使在阅读文档后我也不明白

这是我的数据:

d = {'switchopen': [True, True, False, False, False, False, False, False, False, False, False, False, False, True, True, False, False, True ],
     'switchclose': [ False, False, False, False, False, True, True, True, True, True, False, False, False, False, False, False, False, False]}

df = pd.DataFrame(data=d)

当 switchopen == True & switchclose == False

时打开

当switchopen == False & switchclose == True时关闭

当 switchopen == False & switchclose == False 时它会移动

我当前的解决方案是这样的:

dfopen = df[(df.switchclose == False) & (df.switchopen == True)]
dfopen['Open'] = 'open'
dfclose = df[(df.switchclose == True) & (df.switchopen == False)]
dfclose['Moving'] = 'close'
dfmoving = df[(df.switchclose == False) & (df.switchopen == False)]
dfmoving['Close'] = 'moving'

dfs = [dfopen, dfclose, dfmoving]
dfz = pd.concat(dfs, sort= False)
dfz = dfz.sort_index(ascending=True)

dfz['Position'] = dfz['Position'].fillna(dfz['Open'])
dfz['Position'] = dfz['Position'].fillna(dfz['Close'])
print(dfz['Position'])

这给了我我想要的东西，但我正在寻找更好的解决方案，例如:

df['Close'] = (df.switchclose == True & df.switchopen == False)
df['Open'] = (df.switchclose == False & df.switchopen == False)
df['Moving'] = (df.switchclose == False & df.switchopen == False)

然后使用.fillna。

如果有一个解决方案甚至不需要该步骤，那就最好了

预期输出是这样的:

       Position
0       open
1       open
2     moving
3     moving
4     moving
5      close
6      close
7      close
8      close
9      close
10    moving
11    moving
12    moving
13      open
14      open
15    moving
16    moving
17      open

Answer 1

考虑到您的逻辑要求和输出期望，这应该可行

df.loc[(df['switchopen'] == True) & (df['switchclose'] == False), 'Position'] = 'open'
df.loc[(df['switchopen'] == False) & (df['switchclose'] == True), 'Position'] = 'closed'
df.loc[(df['switchopen'] == False) & (df['switchclose'] == False), 'Position'] = 'moving'