Question

我喜欢 df：

col_A
[1,2,3]
[2,3]
[1,3]

和 dict 一样：

dd = {1: "Soccer", 2: "Cricket", 3: "Hockey"}

我如何创建一个新列 col_B 像：

col_A      col_B
[1,2,3]    ["Soccer", "Cricket", "Hockey"]
[2,3]      ["Cricket", "Hockey"]
[1,3]      ["Soccer", "Hockey"]

尝试过类似的东西：

df['sports'] = df['col_A'].map(dd)

出现错误：

TypeError: unhashable type: 'list'

Answer 1

您可以使用带有 if 的列表理解来过滤掉不匹配的值：

df['sports'] = df['col_A'].map(lambda x: [dd[y] for y in x if y in dd])

如果不匹配，则替换为 None：

df['sports'] = df['col_A'].map(lambda x: [dd.get(y, None) for y in x])

如果不匹配则返回相同的值：

df['sports'] = df['col_A'].map(lambda x: [dd.get(y, y) for y in x])

示例：

df['sports1'] = df['col_A'].map(lambda x: [dd[y] for y in x if y in dd])
df['sports2'] = df['col_A'].map(lambda x: [dd.get(y, None) for y in x])
df['sports3'] = df['col_A'].map(lambda x: [dd.get(y, y) for y in x])
print (df)
          col_A                    sports1                          sports2  \
0  [1, 2, 3, 5]  [Soccer, Cricket, Hockey]  [Soccer, Cricket, Hockey, None]   
1        [2, 3]          [Cricket, Hockey]                [Cricket, Hockey]   
2        [1, 3]           [Soccer, Hockey]                 [Soccer, Hockey]   

                        sports3  
0  [Soccer, Cricket, Hockey, 5]  
1             [Cricket, Hockey]  
2              [Soccer, Hockey]

Answer 2

col_A=[[1,2,3],[2,3],[1,3]]

df=pd.DataFrame({'col_A':col_A})

dictLkup = {1: "Soccer", 2: "Cricket", 3: "Hockey"}

def lookupfunc(mylist):
    retlist=[] 
    [retlist.append(dictLkup[x]) for x in mylist]
    return(retlist)

df['col_B']=df['col_A'].apply(lambda x: lookupfunc(x))

 print(df.head())

熊猫映射列表到新列的字典项

2 个答案: