Python：如何基于给定文件夹的内容创建具有名称的列表？

Question

I need to create a list like this: myindex=['event 1','event 2','event 3', ..., 'event n'] , where n is the number of csv files I have in a given directory.

To implement this I wrote the following:

directoryPath=raw_input('Directory for csv files: ')
myindex=[] #'event 1','event 2','event 3','event 4','event 5','event 6'
for i,file in enumerate(os.listdir(directoryPath)):
    if file.endswith(".csv"):
        myindex.append('event '+str(i))

The directory I pass it has n csv files and n jpg files, but I only want to consider the former.

The result I get: myindex=['event 0','event 2', 'event 4', ..., 'event 2(n-1)'] .

The result I want: myindex=['event 1', 'event 2', 'event 3', 'event 4', ..., 'event n']

What should I amend in my block? Thanks!

EDIT

What if I want the first element of myindex to be an empty string ? So to have myindex=['','event 1','event 2',...,'event n'] .

Answer 1

enumerate takes a second parameter that is the number to start with. Use 1 . Then use glob to count only csv files:

for i,v in enumerate(glob.glob('*.csv'),1):

Answer 2

Use a genex to filter the filenames.

enumerate((x for os.listdir(directoryPath) if x.endswith(".csv")), start=1)

Or have the OS do it.

enumerate(glob.glob(os.path.join(directoryPath, "*.csv")), start=1)

Answer 3

You need a counter instead of using enumerate since that increments for every loop

directoryPath=raw_input('Directory for csv files: ')
myindex=[''] #'event 1','event 2','event 3','event 4','event 5','event 6'

count = 0
for file in os.listdir(directoryPath):
    if file.endswith(".csv"):
        count += 1
        myindex.append('event '+str(count))