Question

我的数据框为：

<table>
    <tr>
        <td><input type="checkbox" name="sd3" value="mfi_nam9" class="checkme" /></td>
        <td>First Value</td>
        <td><input type="number" name="mfi_nam9" class="text1 required" hidden></td>
        <td>
            <select name="ItemCode" class="text" hidden>
                <option value="001">Item 1</option>
                <option value="002">Item 2</option>
            </select>
        </td>
    </tr>
    <tr>
        <td><input type="checkbox" name="sd3" value="mfi_nam9" class="checkme" /></td>
        <td>First Value</td>
        <td><input type="number" name="mfi_nam9" class="text1 required" hidden></td>
        <td>
            <select name="ItemCode" class="text" hidden>
                <option value="001">Item 1</option>
                <option value="002">Item 2</option>
            </select>
        </td>
    </tr>
</table>



<script>
    document.querySelectorAll('.checkme[type="checkbox"]').forEach(function (elem) {
        elem.addEventListener('change', function () {
            this.parentNode.parentNode.querySelector('.text1[type="number"]').hidden = !this.checked;
            this.parentNode.parentNode.querySelector('select.text').hidden = !this.checked;
        });
    });
</script>

我的字典为：

df = pd.DataFrame(
    {'title':['a1','a2','a3','a4','a5'],
     'genre_name':[
         ['family', 'animation'],
         ['action', 'family', 'comedy'],
         ['family', 'comedy'],
         ['horror','action'],
         ['family', 'animation','comedy']]}
)

df
      title    genre_name
0      a1     ['family', 'animation']
1      a2     ['action', 'family', 'comedy']
2      a3     ['family', 'comedy']
3      a4     ['horror','action]
4      a5     ['family', 'animation','comedy']

我想创建一个称为'genre_ids'的新列，它将所有genre_names映射到字典'dict'中的键。

所需的df为：

dict={'1':'family','2':'animation','3':'action','4':'comedy','5':'horror'}

我该如何实现？

Answer 1

将字典名称从dict更改为另一个变量，因为内建函数（Python代码字），然后将键与列表推导中的值和映射值交换：

d={'1':'family','2':'animation','3':'action','4':'comedy','5':'horror'}

d1 = {v:k for k, v in d.items()}
df['genre_ids'] = df['genre_name'].apply(lambda x: [d1.get(y) for y in x])
#alternative
#df['genre_ids'] = [[d1.get(y) for y in x] for x in df['genre_name']]
print (df)
  title                   genre_name  genre_ids
0    a1          [family, animation]     [1, 2]
1    a2     [action, family, comedy]  [3, 1, 4]
2    a3             [family, comedy]     [1, 4]
3    a4             [horror, action]     [5, 3]
4    a5  [family, animation, comedy]  [1, 2, 4]

编辑：您还可以指定如果不匹配将发生什么情况，在这里为第一个列表添加crime：

df = pd.DataFrame({'title':['a1','a2','a3','a4','a5'], 
                   'genre_name':[['crime', 'animation'],['action', 'family', 'comedy'],
                                 ['family', 'comedy'],['horror','action'],
                                 ['family', 'animation','comedy']]})

d={'1':'family','2':'animation','3':'action','4':'comedy','5':'horror'}


d1 = {v:k for k, v in d.items()}
#no matched values repalced to None
df['genre_ids0'] = df['genre_name'].apply(lambda x: [d1.get(y) for y in x])
#no match replaced to default value
df['genre_ids1'] = df['genre_name'].apply(lambda x: [d1.get(y, 0) for y in x])
#no match is removed
df['genre_ids2'] = df['genre_name'].apply(lambda x: [d1[y] for y in x if y in d1])
print (df)
  title                   genre_name genre_ids0 genre_ids1 genre_ids2
0    a1           [crime, animation]  [None, 2]     [0, 2]        [2]
1    a2     [action, family, comedy]  [3, 1, 4]  [3, 1, 4]  [3, 1, 4]
2    a3             [family, comedy]     [1, 4]     [1, 4]     [1, 4]
3    a4             [horror, action]     [5, 3]     [5, 3]     [5, 3]
4    a5  [family, animation, comedy]  [1, 2, 4]  [1, 2, 4]  [1, 2, 4]

如何将字典值映射到具有值列表的数据框列

1 个答案: