To create a list of all file name & it's data. The first element of list should be the file nam-CodePudding

Note, the first element of the list should be the filename and second element should be it's corresponding data.

And I've to do this for all the files in the file_path

for i in file_path:
    txt_file=open(i, 'r')
    data = txt_file.read()
    k=0 
    filename[k] = filename[k] ':' data
    k =1
        
    txt_file.close()

filename = ['PATIENT#3 PAGE 1.txt','PATIENT #3 PAGE 2.txt','PATIENT #5 PAGE 1.txt','PATIENT #5 PAGE 2.txt',..]

file_path = ['C:\Users\OCR_output\PATIENT #3 PAGE 1.txt,'C:\Users\OCR_output\PATIENT #3 PAGE 2.txt']

My output needs to look like this: [filename1, 'the text data inside file1', 'filename2', 'the text data inside file2']

Please help me to get the desired output. Thanks!

CodePudding user response：

So first things first: you are updating k in every iteration of the for-loop so it always becomes 0. So you are only updating the first element. We can fix this by simply putting the k outside of the loop:

k=0 
for i in file_path:
    txt_file=open(i, 'r')
    data = txt_file.read()
    filename[k] = filename[k] ':' data
    k =1
        
    txt_file.close()

Now I'm going to show a few improvements to this code, assuming the description you have given. First, let's use a with statement to avoid manually opening and closing the file:

k=0
for i in file_path:
    with open(i, 'r')as txt_file:
        data = txt_file.read()
        filename[k] = filename[k] ':' data
        k =1

furthermore, you can use append to add data to your collection, which I'm assuming you want:

for path in file_path:
    with open(path, 'r')as txt_file:
        data = txt_file.read()
        filename.append(i)
        filename.append(data)
        # or: filename.extend([i, data])