在这里寻找等效的解决方案后，我只是为我的需求做了一个简单的进度类。我想我应该把它贴出来。

from __future__ import print_function
import sys
import re


class ProgressBar(object):
    DEFAULT = 'Progress: %(bar)s %(percent)3d%%'
    FULL = '%(bar)s %(current)d/%(total)d (%(percent)3d%%) %(remaining)d to go'

    def __init__(self, total, width=40, fmt=DEFAULT, symbol='=',
                 output=sys.stderr):
        assert len(symbol) == 1

        self.total = total
        self.width = width
        self.symbol = symbol
        self.output = output
        self.fmt = re.sub(r'(?P<name>%\(.+?\))d',
            r'\g<name>%dd' % len(str(total)), fmt)

        self.current = 0

    def __call__(self):
        percent = self.current / float(self.total)
        size = int(self.width * percent)
        remaining = self.total - self.current
        bar = '[' + self.symbol * size + ' ' * (self.width - size) + ']'

        args = {
            'total': self.total,
            'bar': bar,
            'current': self.current,
            'percent': percent * 100,
            'remaining': remaining
        }
        print('\r' + self.fmt % args, file=self.output, end='')

    def done(self):
        self.current = self.total
        self()
        print('', file=self.output)

例子:

from time import sleep

progress = ProgressBar(80, fmt=ProgressBar.FULL)

for x in xrange(progress.total):
    progress.current += 1
    progress()
    sleep(0.1)
progress.done()

将打印以下内容:

[========] 17/80 (21%) 63

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")

有一些特定的库(比如这里的这个)，但也许一些非常简单的库就可以做到:

import time
import sys

toolbar_width = 40

# setup toolbar
sys.stdout.write("[%s]" % (" " * toolbar_width))
sys.stdout.flush()
sys.stdout.write("\b" * (toolbar_width+1)) # return to start of line, after '['

for i in range(toolbar_width):
    time.sleep(0.1) # do real work here
    # update the bar
    sys.stdout.write("-")
    sys.stdout.flush()

sys.stdout.write("]\n") # this ends the progress bar

注意:progressbar2是progressbar的一个分支，已经很多年没有维护过了。

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%

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)

受到许多不依赖包的答案的启发，我在这里分享我的实现。在任何循环中使用的函数都需要当前迭代数、迭代总数和初始时间。

import time    
def simple_progress_bar(i: int, n: int, init_time: float):
    avg_time = (time.time()-init_time)/(i+1)
    percent = ((i+1)/(n))*100
    print(
        end=f"\r|{'='*(int(percent))+'>'+'.'*int(100-int(percent))}|| " + \
        f"||Completion: {percent : 4.3f}% || \t "+ \
        f"||Time elapsed: {avg_time*(i+1):4.3f} seconds || \t " + \
        f"||Remaining time: {(avg_time*(n-(i+1))): 4.3f} seconds."
    )
    return



N = 325
t0 = time.time()
for k in range(N):
    # stuff goes here #
    time.sleep(0.0001)
    # stuff goes here #
    
    simple_progress_bar(k, N, t0)