如果它是一个大循环，迭代次数固定，需要花费很多时间，你可以使用我做的这个函数。循环的每一次迭代都增加了进展。其中count是循环的当前迭代，total是循环的值，size(int)是你想要的条的大小，以10为增量，即(size 1 =10个字符，size 2 =20个字符)

import sys
def loadingBar(count,total,size):
    percent = float(count)/float(total)*100
    sys.stdout.write("\r" + str(int(count)).rjust(3,'0')+"/"+str(int(total)).rjust(3,'0') + ' [' + '='*int(percent/10)*size + ' '*(10-int(percent/10))*size + ']')

例子:

for i in range(0,100):
     loadingBar(i,100,2)
     #do some code 

输出:

i = 50
>> 050/100 [==========          ]

一个非常简单的方法:

def progbar(count: int) -> None:
    for i in range(count):
        print(f"[{i*'#'}{(count-1-i)*' '}] - {i+1}/{count}", end="\r")
        yield i
    print('\n')

以及用法:

from time import sleep

for i in progbar(10):
    sleep(0.2) #whatever task you need to do

这是我的简单解决方案:

import time

def progress(_cur, _max):
    p = round(100*_cur/_max)
    b = f"Progress: {p}% - ["+"."*int(p/5)+" "*(20-int(p/5))+"]"
    print(b, end="\r")

# USAGE:
for i in range(0,101):
    time.sleep(0.1) 
    progress(i,100)

print("..."*5, end="\r")
print("Done")

我使用format()方法来制作一个加载条。以下是我的解决方案:

import time

loadbarwidth = 23

for i in range(1, loadbarwidth + 1):
    time.sleep(0.1) 

    strbarwidth = '[{}{}] - {}\r'.format(
        (i * '#'),
        ((loadbarwidth - i) * '-'),
        (('{:0.2f}'.format(((i) * (100/loadbarwidth))) + '%'))
    )

    print(strbarwidth ,end = '')

print()

输出:

[#######################] - 100.00%

我喜欢Brian Khuu的答案，因为它简单，不需要外部包。我做了一点改动，所以我在这里添加了我的版本:

import sys
import time


def updt(total, progress):
    """
    Displays or updates a console progress bar.

    Original source: https://stackoverflow.com/a/15860757/1391441
    """
    barLength, status = 20, ""
    progress = float(progress) / float(total)
    if progress >= 1.:
        progress, status = 1, "\r\n"
    block = int(round(barLength * progress))
    text = "\r[{}] {:.0f}% {}".format(
        "#" * block + "-" * (barLength - block), round(progress * 100, 0),
        status)
    sys.stdout.write(text)
    sys.stdout.flush()


runs = 300
for run_num in range(runs):
    time.sleep(.1)
    updt(runs, run_num + 1)

它取总运行次数(total)和目前处理的运行次数(progress)，假设total >= progress。结果如下所示:

[#####---------------] 27%

这里有一个简单的解决方案!

void = '-'
fill = '#'
count = 100/length
increaseCount = 0
for i in range(length):
    print('['+(fill*i)+(void*(length-i))+'] '+str(int(increaseCount))+'%',end='\r')
    increaseCount += count
    time.sleep(0.1)
print('['+(fill*(i+1))+(void*(length-(i+1)))+'] '+str(int(increaseCount))+'%',end='\n')

注意:如果需要，可以修改fill和“void”字符。

加载条(图片)