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首页 > 解决方案 > 如何使用预定义列表中的单词出现在数据框的文本列中的次数的计数器创建一个新列？

python - 如何使用预定义列表中的单词出现在数据框的文本列中的次数的计数器创建一个新列？

问题描述

我想构建一个新列，其中包含 ai_functional 列表中的单词在文本列中出现的次数的计数。

给出的清单是：

> ai_functional = ["natural language
> processing","nlp","A I ","Aritificial intelligence", "stemming","lemmatization","lemmatization","information
> extraction","text mining","text analytics","data-mining"]

我想要的结果如下：

> text                                                                     counter
> 
> 1. More details  A I   Artificial Intelligence                             2
> 2. NLP works very well these days                                          1         
> 3. receiving information at the right time                                 1

我一直在使用的代码是

def func(stringans):
  for x in ai_tech:
    count = stringans.count(x)
  
  return count

df['counter']=df['text'].apply(func)

请有人可以帮我解决这个问题。我真的被卡住了，因为每次我应用这个我都会在计数器列中得到结果为 0

标签： pythonpandasdataframecounterdata-manipulation

解决方案

当你这样做count = 时，你删除了以前的值，你想总结不同的计数

def func(stringans):
    count = 0
    for x in ai_tech:
        count += stringans.count(x)
    return count

# with sum and generator 
def func(stringans):
    return sum(stringans.count(x) for x in ai_tech)

修复了一些拼写错误ai_tech并将所有设置.lower()为计数器 col 中的给定2,1,0，最后一行没有共同值

import pandas as pd

ai_tech = ["natural language processing", "nlp", "A I ", "Artificial intelligence",
           "stemming", "lemmatization", "information extraction",
           "text mining", "text analytics", "data - mining"]

df = pd.DataFrame([["1. More details  A I   Artificial Intelligence"], ["2. NLP works very well these days"],
                   ["3. receiving information at the right time"]], columns=["text"])

def func(stringans):
    return sum(stringans.lower().count(x.lower()) for x in ai_tech)

df['counter'] = df['text'].apply(func)
print(df)

# ------------------
                                             text  counter
0  1. More details  A I   Artificial Intelligence        2
1               2. NLP works very well these days        1
2      3. receiving information at the right time        0

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