如何通过将列表元素与 Python 中的字符串进行比较来找到列表元素的索引？

Question

Suppose if I had a list that contains some files, like so:

file_list = ['folder\\text.txt', 'folder\\data.csv', 'folder\\picture.png']

And also declared a string:

file = 'text.txt'  #Did not include 'folder\\', intentionally.

How can I grab the index of the file_list by using the 'file' variable?

I have tried the following:

index = file_list.index(file)
filename = file_list[index]
print(filename)

However, I am getting this result:

ValueError: 'text.txt' is not in list

I've understood why my solution failed but is there an efficient solution for grabbing the index by using the string variable?

Answer 1

This will help you:)

file_list = ['folder\\text.txt', 'folder\\data.csv', 'folder\\picture.png']
file = 'data.csv'
flag = True
for i in range(len(file_list)):
    if(file_list[i].find(file)!=-1):
        print("file found at {} position in the list".format(i))
        flag = False
        break
if flag:
    print ("File not found :(")

OUTPUT:

file found at 0 position in the list

Answer 2

given you don't want to use os module (as you said in the comment), just implement a basic basename func based on split.

then, you can use a simple comprehension to create a matching list with just the filenames, and look for the index in it.

def basename(path):
  return path.split('\\')[-1]

file_list = ['folder\\text.txt', 'folder\\data.csv', 'folder\\picture.png']
file = 'text.txt'

file_basename_list = [basename(f) for f in file_list]

index = file_basename_list.index(file)

filename = file_list[index]
print(filename)

Output:

folder\text.txt

Answer 3

You should check each element of file_list to findout the index of file :

def find_index(file, file_list):
    for idx, file_name in enumerate(file_list):
        if file in file_name:
            print(f"idx:{idx}, file_name:{file_name}")
            return idx


if __name__ == '__main__':
    file_list = ['folder\\text.txt', 'folder\\data.csv', 'folder\\picture.png']
    file = 'text.txt'
    idx = find_index(file, file_list)
    if idx:
        print(file_list[idx])
    else:
        print(f"Couldn't find idx of {file} ")

Output:

idx:0, file_name:folder\text.txt
folder\text.txt

Answer 4

The straightforward way to do this would be:

file_list = ['folder\\text.txt', 'folder\\data.csv', 'folder\\picture.png']
search_file = 'text.txt'
base_path = "folder\\"

def find_file(file_list, search_file, base_path):
    for idx, file in enumerate(file_list):
        if file == base_path+search_file:
            return idx
    return "File Not Found"

print(find_file(file_list, search_file, base_path))

If you want to add the base_path dynamically then use:

import os.path 

path = 'folder\\'
base_path = os.path.basename(path)