Question

我正在尝试更改大型数据框中的元素

e.g

Keyword                         |cat1|cat2|cat3|
--------------------------------|----|----|----|
beach holiday                   |    |    |    |
package beach holiday           |    |    |    |
inclusive package beach holiday |    |    |    |

我运行一个方法find_keywords（Keyword），该方法传入关键字，例如“包容性海滩假期”，与文本类别列表进行比较，并返回前三个相关类别。

''' 
Input a Keyword, breaks it down and finds which category it matches
'''
def find_keywords(keywords):
words = keywords.split()
wordlist = []
for word in words:      
    if word in categories:
        wordlist.append(word)       
wordlist = wordlist [:3]    
return wordlist

在这种情况下：

['inclusive','package','beach']

这一切都很好，当我在数据上运行我的主方法时

if __name__ == '__main__':


df = get_csv(csv)


for index, row in df.iterrows():
    row['Keyword'].lower()
    print(row['Keyword'])
    tokens = find_keywords(row['Keyword'])
    print(tokens)

它返回：

beach holiday
['beach','holiday']                   
package beach holiday  
['package','beach','holiday']         
inclusive package beach holiday 
['inclusive','package','beach']

我如何获取每个列表并将其添加到cat1 / cat2 / cat3列

生成数据框：

Keyword                         |cat1   |cat2    |cat3   |
--------------------------------|----   |----    |----   |
beach holiday                   |beach  |holiday |       |
package beach holiday           |package|beach   |holiday|
inclusive package beach holiday |inclusive|package|beach |

使用@DaFanat的解决方案我能够得到我所要求的但是我对此有轻微的排列，是否可以检查字典而不是列表？

e.g

{'beach': ['beach', 'sand', 'coast'],
'hotel': ['hotel', 'resort']}

然后将头部术语应用于该类别，例如，如果它找到沙子就会将其标记为海滩。

我的尝试：如果名称 =='主要'：

df = get_csv(csv)
h = open('head_categories.txt','r')
mydict = h.read()
mydict = ast.literal_eval(mydict)




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')

Answer 1

我认为这可以胜任：

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

使用iterrows（）

1 个答案: