如何用dict \ df在一列上用条件替换列列表

Posted 2023-03-11

【中文标题】如何用dict \\ df在一列上用条件替换列列表

【英文标题】：how to replace list of columns with condition on one col by dict\df如何用dict \ df在一列上用条件替换列列表

【发布时间】：2020-12-07 09:39:09

【问题描述】：

我喜欢在一个列上按条件更改列列表中的值，使用 dict\df 我得到了 df\dict

类似的东西

list_of_cols=["col1",.."col5"]
df[list_of_cols]=where[df["A"]==dict],dict

这是数据

data="col1":[np.nan,3,4,5,9,2,6],
"col2":[4,2,4,6,0,1,5],
"col3":[7,6,0,11,3,6,7],
"col4":[14,11,22,8,6,np.nan,9],
"col5":[0,5,7,3,8,2,9],
"type":["A","A","C","A","B","A","E"],
"number":["one","two","two","one","one","two","two"]
df=pd.DataFrame.from_dict(data)
df

这是我想将 df["type"] 与 dict 映射的 dict\df 所以 type==A ， col1-col5 将是 0

my_dict="A":0,"B":21,"C":14,"D":9
my_dict=pd.DataFrame.from_dict(my_dict, orient='index')
my_dict

这就是我想要得到的

data="col1":[0,0,14,0,21,0,6],
      "col2":[0,0,14,0,21,0,5],
      "col3":[0,0,14,0,21,0,7],
      "col4":[0,0,14,0,21,0,0.9],
      "col5":[0,0,14,0,21,0,9],
      "type":["A","A","C","A","B","A","E"],
"number":["one","two","two","one","one","two","two"]
df=pd.DataFrame.from_dict(data)
df

【参考方案1】：

IIUC，使用isin过滤掉类型，然后使用apply和map直接赋值：

m = df["type"].isin(my_dict)

df.loc[m, "col1":"col5"] = df.loc[m, "col1":"col5"].apply(lambda d: pd.Series.map(df["type"], my_dict))

print (df)

   col1  col2  col3  col4  col5 type number
0   0.0   0.0   0.0   0.0   0.0    A    one
1   0.0   0.0   0.0   0.0   0.0    A    two
2  14.0  14.0  14.0  14.0  14.0    C    two
3   0.0   0.0   0.0   0.0   0.0    A    one
4  21.0  21.0  21.0  21.0  21.0    B    one
5   0.0   0.0   0.0   0.0   0.0    A    two
6   6.0   5.0   7.0   9.0   9.0    E    two

【讨论】：

它的工作，但如果我的字典是 "A":"col1":0,"col2":1,"col3":2,"col4":1,"col5":2 你会怎么做呢？

使用pd.DataFrame("A":"col1":0,"col2":1,"col3":2,"col4":1,"col5":2).T.combine_first(df.set_index("type"))。