如何在Python中从列表和返回结果列表（元素没有与其他元素的长度相似的长度）中匹配元素的索引？
我有一个很大的TXT文件。我想阅读文件的每一行，并根据文件的第三元素获取输入（列表）。
因此，我将文本文件转换为列表并遍历列表的每个元素以获取所需的输出。
这是我的txt文件的一部分：
34566---There was no file in there---Mr. [email protected]
36122---I found the file in [email protected]
64322
28890---I went to see the crowd---Henry [email protected]
44533---The weather made it perfect---Merry [email protected]
这是我使用的代码：
value = input("Enter your name:")
filename = open("test.txt", "r")
for line in filename:
line_num = line.strip('\n').split("---")
if line_num [2] == value:
print(line_num[1])
break
唯一的问题是中间的值，并不是列表中的其他元素。因此，它给出了一个错误，
IndexError: list index out of range
我不知道如何编码如何忽略中间元素，而中间元素并非列表中的其余元素。请帮我。
I have a very big txt file. I want to read through each line of the file and get an input based on the 3rd element of the file(list).
So, I convert the text file to list and iterate through each element of the list to get my desired output.
This is a part of my txt file:
34566---There was no file in there---Mr. [email protected]
36122---I found the file in [email protected]
64322
28890---I went to see the crowd---Henry [email protected]
44533---The weather made it perfect---Merry [email protected]
This is the code that I used:
value = input("Enter your name:")
filename = open("test.txt", "r")
for line in filename:
line_num = line.strip('\n').split("---")
if line_num [2] == value:
print(line_num[1])
break
The only problem is the value in the middle doesnot behave as the other elements in the list. So, it is giving an error
IndexError: list index out of range
I don't know how to code to ignore the middle element which doesnot behave as the rest of the elements in the list. Please help me.
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您可以尝试一下，
如您所见
You can try this,
As You can see I added only the
if line == '\n': continuepart which basically skips the empty line将您的病情更改为：
Change your condition to: