我正在编写这个程序,它使用线程同时下载多个文件,并报告屏幕上每个文件的进度。 我正在使用一个不同的线程来下载每个文件。这没问题:
for file_url, filename in zip(file_urls, filenames):
th = Thread(target=download_file, args=(file_url, filename))
threads.append(th)
th.start()
for thread in threads:
thread.join()
问题是我不知道我可以在屏幕上打印每个文件进度报告,如下所示:
"Downloading 'example1.zip': 54366 bytes of 2240799 70.7%"
"Downloading 'example2.zip': 31712 bytes of 1924639 64.7%"
"Downloading 'example3.zip': 21712 bytes of 3224979 34.7%"
以下代码段用于单行进度报告:
def chunk_report(bytes_so_far, total_size, filename):
percent = float(bytes_so_far) / total_size
percent = round(percent * 100, 2)
print "Downloading '{0}': {1} of {2} {3:3.2g}% \r".format(filename,
bytes_so_far, total_size, percent),
,输出如下:
"Downloading 'example2.zip': 31712 bytes of 1924639 64.7%"
每次线程调用此函数时,它都会更新线程正在下载的文件的屏幕。
所以,问题是如何打印多线进度报告,就像我在python中说明的那样?
提前致谢。
答案 0 :(得分:0)
我会使用Queue向报告主题报告进度:
put
进度消息到队列模拟示例:
import threading
import time
import random
import Queue
import sys
# a downloading thread
def worker(path, total, q):
size = 0
while size < total:
dt = random.randint(1,3)
time.sleep(dt)
ds = random.randint(1,5)
size = size + ds
if size > total: size = total
q.put(("update", path, total, size))
q.put(("done", path))
# the reporting thread
def reporter(q, nworkers):
status = {}
while nworkers > 0:
msg = q.get()
if msg[0] == "update":
path, total, size = msg[1:]
status[path] = (total, size)
# update the screen here
show_progress(status)
elif msg[0] == "done":
nworkers = nworkers - 1
print ""
def show_progress(status):
line = ""
for path in status:
(total, size) = status[path]
line = line + "%s: %3d/%d " % (path, size,total)
sys.stdout.write("\r"+line)
sys.stdout.flush()
def main():
q = Queue.Queue()
w1 = threading.Thread(target = worker, args = ("abc", 30, q) )
w2 = threading.Thread(target = worker, args = ("foobar", 25, q))
w3 = threading.Thread(target = worker, args = ("bazquux", 16, q))
r = threading.Thread(target = reporter, args = (q, 3))
for t in [w1,w2,w3,r]: t.start()
for t in [w1,w2,w3,r]: t.join()
main()