Python counts the number of all files under the specified folder
I have always had this requirement, but I used to write it recursively, feeling unfriendly to the occupation of resources, and python’s maximum recursion depth is not more than 1,000, so I changed it and implemented it by breadth-first traversal.
A total of 24 folders on two floors and a total of 50w of documents were measured. The running time is about 3 seconds. The following is the code:
import os
import queue
def get_file_quantity(folder: str) -> int:
'''BFS获取文件夹下文件的总数量'''
# 判断初始文件夹
assert os.path.isdir(folder), '请输入有效的文件夹参数'
file_quantity = 0 # 初始化文件数量
folder_path_queue = queue.Queue()
folder_path_queue.put_nowait(folder) # 初始化队列的值
# 处理队列里的文件夹
while not folder_path_queue.empty():
folder = folder_path_queue.get_nowait()
file_folder_list = list(map(lambda bar: os.path.join(folder, bar), os.listdir(folder)))
folder_list = list(filter(lambda bar: os.path.isdir(bar), file_folder_list))
for folder_path in folder_list:
folder_path_queue.put_nowait(folder_path)
temp_file_count = len(file_folder_list) - len(folder_list)
file_quantity += temp_file_count
return file_quantity
if __name__ == '__main__':
file_quantity = get_file_quantity(r'/home')
print(f'文件总数是: {file_quantity}')
Train of thought
The queue is mainly used here, which is the common idea of BFS.
A little change herefolder_list = list(filter(lambda bar: os.path.isdir(bar), file_folder_list))
From insidelambda
Function can realize various judgment operations on file name, and the realization of function function here depends entirely on one’s own brain hole!
Attach an adapted version: view the number of files that contain a specific suffix
Adapted version: view the number of files with specific suffixes
import os
import queue
def filter_extension(filename: str, extension: str) -> bool:
'''判断文件路径名的后缀是否和给定的后缀字符串相同
只是单纯的字符串判断
'''
basename_and_extension = filename.split('.')
return (basename_and_extension[-1] == extension) and (len(basename_and_extension) >= 2)
def get_file_quantity(folder: str, extension: str) -> int:
'''BFS获取文件夹下文件的总数量'''
# 判断初始文件夹
assert os.path.isdir(folder), '请输入有效的文件夹参数'
assert isinstance(extension, str), '请输入有效的文件后缀名'
file_quantity = 0 # 初始化文件数量
folder_path_queue = queue.Queue()
folder_path_queue.put_nowait(folder) # 初始化队列的值
# 处理队列里的文件夹
while not folder_path_queue.empty():
folder = folder_path_queue.get_nowait()
file_folder_list = list(map(lambda bar: os.path.join(folder, bar), os.listdir(folder)))
folder_list = list(filter(lambda bar: os.path.isdir(bar), file_folder_list))
file_list = list(filter(lambda bar: os.path.isfile(bar), file_folder_list))
match_extension_list = list(filter(lambda bar: filter_extension(bar, extension), file_list))
for folder_path in folder_list:
folder_path_queue.put_nowait(folder_path)
temp_file_count = len(match_extension_list)
file_quantity += temp_file_count
return file_quantity
if __name__ == '__main__':
extension = 'py'
file_quantity = get_file_quantity(r'/home', extension)
print(f'包含后缀 {extension } 的文件的数量: {file_quantity}')