Question

我正在尝试使用字典对列表中的元素进行分类：

mydict = {'beach': ['beach', 'sand', 'coast'], 'package': ['package', 'inclusive']}

给出数据框：

Keyword                         |cat |
--------------------------------|----|
beach holiday                   |    |
package beach holiday           |    |
inclusive beach holiday         |    |

我想检查一个元素是否在字典中，它将关键字应用于类别列，例如：

Keyword                         |cat |
--------------------------------|----|
beach holiday                   |beach  |
package beach holiday           |package|
inclusive package beach holiday |package|

我尝试使用以下代码：

df = get_csv(csv)
mydict = {'beach': ['beach', 'sand', 'coast'], 'package': ['package', 'inclusive']}

for key in mydict.keys():
    item = key
    if item in mydict[key]:
        target_cats = item
        find_keywords = lambda kw: [s for s in kw.split() if s in target_cats]

        df.loc[:, 'cat_list'] = df['Keyword'].apply(lambda x: find_keywords(x))
        for i in range(1, 4):
            df.loc[:, 'cat{0}'.format(i)] = df['cat_list'].apply(lambda x: x[i-1] if len(x) >= i else '')

print(df)
df.to_csv('kuoniTesting.csv')

然而，这只是一个空的类别列表，检查列表的代码是否有效，如何修改它以使用字典？

target_cats = ['cat', 'dog', 'cow']
df = pd.DataFrame({'Keyword': ['cat dog cow', 'cat dog', 'dog sheep']})
find_keywords = lambda kw: [s for s in kw.split() if s in target_cats]

df.loc[:, 'cat_list'] = df['Keyword'].apply(lambda x: find_keywords(x))
for i in range(1, 4):
    df.loc[:, 'cat{0}'.format(i)] = df['cat_list'].apply(lambda x: x[i-1] if 
    len(x) >= i else '')

 Keyword      cat_list         cat1 cat2 cat3
  0  cat dog cow  [cat, dog, cow]  cat  dog  cow
  1      cat dog       [cat, dog]  cat  dog     
  2    dog sheep            [dog]  dog

Answer 1

您可以使用键在字典中交换values，但它会返回多个值：

mydict = {'beach': ['beach', 'sand', 'coast'], 'package': ['package', 'inclusive']}
d = {k: oldk for oldk, oldv in mydict.items() for k in oldv}
print (d)
{'sand': 'beach', 'package': 'package', 'beach': 'beach', 
 'inclusive': 'package', 'coast': 'beach'}

find_keywords = lambda kw: [d[s] for s in kw.split() if s in d.keys()]
df['cat_list'] = df['Keyword'].apply(lambda x: find_keywords(x))
print (df)
                   Keyword          cat_list
0            beach holiday           [beach]
1    package beach holiday  [package, beach]
2  inclusive beach holiday  [package, beach]

对于新列：

df = df.join(pd.DataFrame(df['cat_list'].values.tolist(), columns=['cat1','cat2']))
print (df)
                   Keyword          cat_list     cat1   cat2
0            beach holiday           [beach]    beach   None
1    package beach holiday  [package, beach]  package  beach
2  inclusive beach holiday  [package, beach]  package  beach

通过检查字典来修改数据框中的元素

1 个答案: