Python:列出目录及其大小
我正在尝试写一些Python代码,目的是遍历当前工作目录下的每个文件夹,并报告每个文件夹下的总大小(以字节为单位),不管这些文件夹有多深。
这只是一个学习项目,我知道通过命令行已经有其他方法可以获取这些信息。以下是我目前写的代码:
# get name of current working directory
start_directory = os.getcwd()
# create dictionary to hold the size of each folder in
# the current working directory
top_level_directory_sizes = {}
# initialize directory
for i in os.listdir(start_directory):
if os.path.isdir(i):
top_level_directory_sizes[i] = 0
# traverse all paths from current working directory
for dirpath, dirnames, filenames in os.walk(start_directory):
for f in filenames:
fp = os.path.join(dirpath, f)
#increment appropriate dictionary element: += os.path.getsize(fp)
for k,v in top_level_directory_sizes.iteritems():
print k, v
所以希望输出的结果看起来像这样:
algorithms 23,754 bytes
articles 1,234 bytes
books 123,232 bytes
images 78,232 bytes
total 226,452 bytes
2 个回答
2
你可以看看 os.path.walk
这个东西。
4
这段代码会列出一个指定文件夹里所有子文件夹的大小,以及它们的总大小:
import locale
import os
locale.setlocale(locale.LC_ALL, "")
def get_size(state, root, names):
paths = [os.path.realpath(os.path.join(root, n)) for n in names]
# handles dangling symlinks
state[0] += sum(os.stat(p).st_size for p in paths if os.path.exists(p))
def print_sizes(root):
total = 0
paths = []
state = [0]
n_ind = s_ind = 0
for name in sorted(os.listdir(root)):
path = os.path.join(root, name)
if not os.path.isdir(path):
continue
state[0] = 0
os.path.walk(path, get_size, state)
total += state[0]
s_size = locale.format('%8.0f', state[0], 3)
n_ind = max(n_ind, len(name), 5)
s_ind = max(s_ind, len(s_size))
paths.append((name, s_size))
for name, size in paths:
print name.ljust(n_ind), size.rjust(s_ind), 'bytes'
s_total = locale.format('%8.0f', total, 3)
print '\ntotal'.ljust(n_ind), s_total.rjust(s_ind), 'bytes'
print_sizes('.')
输出结果:
% python dirsizes.py
bar 102,672 bytes
foo 102,400 bytes
total 205,072 bytes