python中与字典键匹配的字符串

Question

我有一个字符串列表和一本字典。例如：

list = ["apple fell on Newton", "lemon is yellow","grass is greener"]
dict = {"apple" : "fruits", "lemon" : "vegetable"}

任务是将列表中的每个字符串与字典的键相匹配。如果匹配，则返回键的值

目前，我正在使用这种非常耗时的方法。有人能帮我找到一些有效的方法吗

lmb_extract_type = (lambda post: list(filter(None, set(dict.get(w)[0] if w in post.lower().split() else None for w in dict))))

 df['type']  = df[list].apply(lmb_extract_type)

Answer 1

It is a single column with a string (eg.: "apple fell on Newton") in each row of the data frame. For each row, I have to match it with key from the dictionary and return value of the corresponding key

Number of elements in the list is around 40-50 million.So, its taking a lot of time

IIUC，根据您的评论，您可以使用str.extract和series.replace轻松解决这个问题，这两个函数都是矢量化函数，没有任何循环

1. 对于使用str.extract，可以从字典的键创建正则表达式模式。这只提取关键词苹果或柠檬
2. 您可以使用字典d直接用相应的值替换其中的每一个

l = ["apple fell on Newton", "lemon is yellow","grass is greener"]
d = {"apple" : "fruits", "lemon" : "vegetable"}

df = pd.DataFrame(l, columns=['sentences']) #Single column dataframe to demonstrate.

pattern = '('+'|'.join(d.keys())+')'   #Regular expression pattern
df['type'] = df.sentences.str.extract(pattern).replace(d)
print(df)

              sentences       type
0  apple fell on Newton     fruits
1       lemon is yellow  vegetable
2      grass is greener        NaN