Question

我有一个数据框和一本字典：

news = {'Text':['dog ate the apple', 'cat ate the carrot', 'dog drank water'], 'Source':['NYT', 'WP', 'Guardian']}
news_df = pd.DataFrame(news)


w = {1:['horse', 'dog'], 2:['apple'], 10: ['water', 'melon', 'liquerice']}

我想创建一个新列news_df ['sum']来查看news_df ['Text']，检查是否有任何词典值可用，并且如果这些列中有1个或多个，则分配按键。我的结果将是：

results = {'Text':['dog ate the apple', 'cat ate the carrot', 'dog drank water'], 'Source':['NYT', 'WP', 'Guardian'], 'sum' : [3, 0, 11]}
results_df = pd.DataFrame(results)

有什么想法吗？我不确定采取什么方法？也许把字典变成一个数据框？

Answer 1

这是一种适用方法：

def counts(x): 
   sumcount = 0 
   for k, v in w.items(): 
      for word in v: 
        if word in x: 
           sumcount+=int(k) 
   return sumcount 

news_df.Text.apply(counts)

                 Text    Source  sum
0   dog ate the apple       NYT    3
1  cat ate the carrot        WP    0
2     dog drank water  Guardian   11

根据字典键创建具有值的新列

1 个答案: