通过在字符串列中查找确切的单词来创建新列 [英] Creating a new column by finding exact word in a column of strings

问题描述

如果列表中的任何单词与dataframe字符串列完全匹配，我想用1或0创建一个新列.

I want to create a new column with 1 or 0, if any of the words in a list is matched exaclty with the dataframe string column.

list_provided=["mul","the"]
#how my dataframe looks
id  text
a    simultaneous there the
b    simultaneous there
c    mul why

预期输出

id  text                     found
a    simultaneous there the   1
b    simultaneous there       0
c    mul why                  1

第二行被分配为0，因为 "mul"或"the"在字符串列"text"中都不完全匹配

Second row is assigned 0, since either of "mul" or "the" are not exactly matching in the string column "text"

代码尝试到现在为止

#For exact match I am using the below code
data["Found"]=np.where(data["text"].str.contains(r'(?:\s|^)penalidades(?:\s|$)'),1,0)

如何遍历循环以查找提供的单词列表中所有单词的完全匹配?

How can I iterate through a loop to find exact match of all the words in the provided list of words?

修改: 如果我按照Georgey的建议使用str.contains(pattern)，则data ["Found"]的所有行都变为1

If i use str.contains(pattern) as suggested by Georgey, all the rows for data["Found"] becomes 1

data=pd.DataFrame({"id":("a","b","c","d"), "text":("simultaneous there the","simultaneous there","mul why","mul")})
list_of_word=["mul","the"]
pattern = '|'.join(list_of_word)
data["Found"]=np.where(data["text"].str.contains(pattern),1,0)

Output:
id  text                     found
a    simultaneous there the   1
b    simultaneous there       1
c    mul why                  1
d    mul                      1

找到的列中的第二行应为0

The second row in the found column should be 0 here

推荐答案

您可以使用带有生成器表达式的pd.Series.apply和sum来做到这一点:

You can do this with pd.Series.apply and sum with a generator expression:

import pandas as pd

df = pd.DataFrame({'id': ['a', 'b', 'c'],
                   'text': ['simultaneous there the', 'simultaneous there', 'mul why']})

test_set = {'mul', 'the'}

df['found'] = df['text'].apply(lambda x: sum(i in test_set for i in x.split()))

#   id                    text  found
# 0  a  simultaneous there the      1
# 1  b      simultaneous there      0
# 2  c                 mul why      1

---

上面提供了一个 count .如果只需要布尔值，请使用any:

---

The above provides a count. If you just need a Boolean, use any:

df['found'] = df['text'].apply(lambda x: any(i in test_set for i in x.split()))

对于整数表示，请链接.astype(int).

For integer representation, chain .astype(int).

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