下面是一个简短的解决方案，以编程方式构建加载条(您必须决定需要多长时间)。

import time

n = 33  # or however many loading slots you want to have
load = 0.01  # artificial loading time!
loading = '.' * n  # for strings, * is the repeat operator

for i in range(n+1):
    # this loop replaces each dot with a hash!
    print('\r%s Loading at %3d percent!' % (loading, i*100/n), end='')
    loading = loading[:i] + '#' + loading[i+1:]
    time.sleep(load)
    if i==n: print()

使用这个库:fish (GitHub)。

用法:

>>> import fish
>>> while churning:
...     churn_churn()
...     fish.animate()

玩得开心!

已经有很多好的答案，但添加这个特定的基于@HandyGold75的答案，我希望它在特定的上下文中是callabe，有一个初始的msg，加上在结束时的几秒钟的时间反馈。

from time import sleep, time


class ProgressBar:
    def __init__(self, total: float, width: int = 50, msg: str = ""):
    self.total = total
    self.width = width
    self.start: float = time()
    if msg:
        print(f"{msg}")

    def progress(self, progress: float):
        percent = self.width * ((progress) / self.total)
        bar = chr(9608) * int(percent) + "-" * (self.width - int(percent))
        print(
            f"\r|{bar}| {(100/self.width)*percent:.2f}% "
            f"[{progress} of {self.total}]",
            end="\r",
        )

    def __enter__(self):
        return self.progress

    def __exit__(self, type, value, traceback):
        end: float = time()
        print(f"\nFinished after {end - self.start: .3f} seconds.")


# USAGE
total_loops = 150
with ProgressBar(total=total_loops) as progress:
    for i in range(total_loops):
        sleep(0.01)  # Do something usefull here
        progress(i + 1)

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 [==========          ]

已经有很多令人惊叹的答案，但我想分享我对进度条的解决方案。

from time import sleep

def progress_bar(progress: float, total: float, width: int = 25):
    percent = width * ((progress + 1) / total)
    bar = chr(9608) * int(percent) + "-" * (width - int(percent))
    print(f"\r|{bar}| {(100/width)*percent:.2f}%", end="\r")

numbers = range(0, 1000)
numbersLen = len(numbers)
for i in numbers:
    sleep(0.01) # Do something usefull here
    progress_bar(i, numbersLen)

编辑:

如果你正在寻找一个条，调整它是基于终端的宽度和可能的消息在结束，然后这也是工作。请注意，如果终端太窄，消息将消失，因为如果它太宽，竖条将断开。

def progressBar(progress: float, total: float, message: str = ""):
    terminalWidth = get_terminal_size().columns
    width = int(terminalWidth / 4)
    percent = width * ((progress + 1) / total)
    bar = chr(9608) * int(percent) + "-" * (width - int(percent))
    if terminalWidth <= 40:
        message = ""
    else:
        message = message + (" " * (int(terminalWidth / 2) - len(message)))
    print(f"\r|{bar}| {(100/width)*percent:.2f}% " + message, end="\r")

回答没有外部库的简单进度条

import time, sys

def progress(size):
    for item in range(size):
        if(item==0):
            print("[",end="")

        elif(item==size-1):
            print("]",end="\n")

        else:
            #main work goes here
            time.sleep(0.1)
            print("%",end="")
            sys.stdout.flush()

progress(50)