英文:
Can I use .txt filenames "dynamically" as Keys for a python dictionary?
问题
抱歉，我的英语不太好，请谅解。
其次，我能否创建一个字典，使其“动态”地使用我想要读取/打开的.txt文件的“结尾”作为字典的键？
背景是，我做了很多测量，但忘记在.txt文件中添加某个变量。这些.txt文件有时包含超过5000行（例如时间测量）。
实际上，我只忘记了13个不同的值。.txt文件都以某个计划的名称命名，因此结尾"_12"或"_01"将始终是我在脚本中用于计算的相同值。
目前的问题是，我不太清楚如何正确地将键连接到文件名，以及如何将字典“转换”为“适当”的值（出现了TypeError：unsupported operand type(s) for *: 'int' and 'dict'）。尽管我认为后面的问题可以通过创建一个元组来解决？
非常感谢大家的帮助。
我首先尝试创建一个简单的字典，然后使用简单的“J=I*A”，理想情况下，A将是根据文件名从数据框中获取的某个变量的值。
英文:
First of all, sorry for my bad englisch - please bear with me.
Secondly; can I create a dictionary so "dynamically", that I can use the "Endings" of the .txt Files I want to read/open with the script as keys for a dictionary?
The Background is; I've made alot of measurments, yet I forgot to put in a certain Variable in the .txt Files. Those .txt files though do contain sometimes over 5000 lines (Time measurments e.g.).
Yet I've actually only 13 different values I forgot. The .txt Files are all named after a certain plan, so the Ending _12 or _01 will be always the same value I miss for a calculation in the script.
The issue right now is, that I don't exactly know how connect the key to the file name properly and how to "convert" the Dictionary to a "proper" value. (Getting the TypeError: unsupported operand type(s) for *: 'int' and 'dict') - Although I think the later problem could be solved by creating a tuple?
Thank y'all in advance.
I tried first to create a simple dict and then use simple "J=I*A", ideally A would be a certain variable from the dictionary, depending on the Filename the I is from via Dataframe.
答案1
得分: 0
你显然已经在你的程序中打开文件，所以你应该能够根据以下内容来获取你想要的文件值：
filename = "relative_or_abs_path_to_your_file_01.txt"
# 返回一个列表，包含扩展名 ('txt') 作为元素 1，文件名的其余部分作为元素 0
f = filename.rsplit('.', 1)
# 在元素 0 (文件名元素) 上拆分以获取下划线后的值 (返回列表的元素 1)
missing_value = f[0].rsplit('_', 1)[1]
missing_value = int(missing_value) # 转换为整数
print(missing_value)
一旦你有了 missing_value 变量，你可以将它放入字典或任何你需要的地方，以适应你的其余代码。 (或者直接按原样使用它。)
英文:
You're obviously already opening the file(s) in your program, so you should be able to adapt the following to get the value you want from the file in question:
filename = "relative_or_abs_path_to_your_file_01.txt"
# Returns a list with extension ('txt') as element 1, and the rest of the filename as element 0
f = filename.rsplit('.', 1)
# Split on element 0 (the filename element) to get value after the underscore (element 1 of the returned list)
missing_value = f[0].rsplit('_', 1)[1]
missing_value = int(missing_value) # Convert into an integer
print(missing_value)
Once you have the missing_value variable, you can obviously put it into a dict, or wherever you need it, to accommodate the rest of your code. (Or just use it directly as-is.)
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