实现如下所示的状态栏:

[==========                ]  45%
[================          ]  60%
[==========================] 100%

我想把这个打印到标准输出,并保持刷新,而不是打印到另一行。如何做到这一点?


当前回答

你可以使用\r(回车)。演示:

import sys
total = 10000000
point = total / 100
increment = total / 20
for i in xrange(total):
    if(i % (5 * point) == 0):
        sys.stdout.write("\r[" + "=" * (i / increment) +  " " * ((total - i)/ increment) + "]" +  str(i / point) + "%")
        sys.stdout.flush()

其他回答

最简单的还是

import sys
total_records = 1000
for i in range (total_records):
    sys.stdout.write('\rUpdated record: ' + str(i) + ' of ' + str(total_records))
    sys.stdout.flush()

关键是将整数类型转换为字符串。

要成为纯python并且不进行系统调用:

from time import sleep

for i in range(21):
    spaces = " " * (20 - i)
    percentage = 5*i
    print(f"\r[{'='*i}{spaces}]{percentage}%", flush=True, end="")
    sleep(0.25)

正如Mark Rushakoff的解决方案中所描述的,您可以输出回车符sys.stdout.write('\r')来将光标重置到行首。要泛化该解决方案,同时实现Python 3的f- string,您可以使用

from time import sleep
import sys

n_bar = 50
iterable = range(33)  # for demo purposes
n_iter = len(iterable)
for i, item in enumerate(iterable):
    j = (i + 1) / n_iter

    sys.stdout.write('\r')
    sys.stdout.write(f"[{'=' * int(n_bar * j):{n_bar}s}] {int(100 * j)}%")
    sys.stdout.flush()

    sleep(0.05)  
    # do something with <item> here

在这里你可以使用以下代码作为函数:

def drawProgressBar(percent, barLen = 20):
    sys.stdout.write("\r")
    progress = ""
    for i in range(barLen):
        if i < int(barLen * percent):
            progress += "="
        else:
            progress += " "
    sys.stdout.write("[ %s ] %.2f%%" % (progress, percent * 100))
    sys.stdout.flush()

使用.format:

def drawProgressBar(percent, barLen = 20):
    # percent float from 0 to 1. 
    sys.stdout.write("\r")
    sys.stdout.write("[{:<{}}] {:.0f}%".format("=" * int(barLen * percent), barLen, percent * 100))
    sys.stdout.flush()

这是一个非常简单的方法,可以用于任何循环。

#!/usr/bin/python
for i in range(100001):
    s =  ((i/5000)*'#')+str(i)+(' %')
    print ('\r'+s),