试图创建一个单词计数器，但我的语句在 python 中给出了一个类型错误

Question

python 广告的新手试图了解如何解决此问题-

   import pandas as pd
    import pyodbc
    from collections import Counter, defaultdict
    import numpy as np
    conn = pyodbc.connect('Driver={SQL Server};'
                          'Server=10.50.3.110\ds;'
                          'Database=xx;'
                          'UID=xx;'
                          'PWD=xx@#Fxx*xx>m;'
                          )


transcripts = pd.read_sql_query("SELECT * FROM forward.claims", conn)

conn.close()


transcripts['Transcript'] = transcripts['Transcript'].str.lower()

Terms =  [ 
             ['deductible','deductible'], 
             ['deductible','deduct'], 
             ['deductible','deducted'], 
             ['pre-approved','pre approved'], 
             ['pre-approved','pre approve'], 
             ['medical records','medical records'], 
             ['medical records','medical record'],]
Terms = pd.DataFrame(Terms, columns = ['Terms', 'Keyword'])

list_of_results = []
count = defaultdict(lambda: 0)
for index, row in transcripts.iterrows():
    for string_to_search in Terms['Keyword'].to_list():
        if string_to_search in row['Transcript']:
            count[string_to_search] += row['Transcript'].count(string_to_search)
            list_of_results.append((row['InteractionId'],string_to_search))
            

Traceback (most recent call last):

  File "<ipython-input-20-88446cec2859>", line 5, in <module>
    if string_to_search in row['Transcript']:

TypeError: argument of type 'NoneType' is not iterable

transcripts
Out[22]: 
            InteractionId  ...                                         Transcript
0      200323337020200120  ...                                    donald disease 
1      200323337020200120  ...  does the same thing like the 250 well it's a 2...
2      200323337020200120  ...           do you guys have somebody speak spanish 
3      200323337020200129  ...                do you want me to get a translator 

[39024 rows x 5 columns]

为什么会出现这个错误，我该如何解决这个错误？是否有针对此特定类型错误的通用解决方案？另请在上面找到成绩单的代表

Answer 1

您至少有一行 row['Transcript'] 实际上只是 None。您可以通过多种方式之一避免此错误，但最简单的可能是首先在相同的逻辑子句中检查这种情况。
所以不是 -

if string_to_search in row['Transcript']:

尝试-

if row['Transcript'] is not None and string_to_search in row['Transcript']:

Answer 2

正如其他人所说，row['Transcript'] 在某些时候是 None。事先检查此表达式是否为 not None 会有所帮助。如果您正在寻找一个优雅的解决方案，我只是想给您一个提示：collections.Counter()

试图创建一个单词计数器，但我的语句在 python 中给出了一个类型错误

Trying to create a word counter but my statement gives a typeerror in python

python

if-statement

typeerror

pandas