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

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

如果你的工作不能被分解成可测量的块，你可以在一个新的线程中调用你的函数，并记录它所花费的时间:

import thread
import time
import sys

def work():
    time.sleep( 5 )

def locked_call( func, lock ):
    lock.acquire()
    func()
    lock.release()

lock = thread.allocate_lock()
thread.start_new_thread( locked_call, ( work, lock, ) )

# This part is icky...
while( not lock.locked() ):
    time.sleep( 0.1 )

while( lock.locked() ):
    sys.stdout.write( "*" )
    sys.stdout.flush()
    time.sleep( 1 )
print "\nWork Done"

显然，您可以根据需要提高计时精度。

我喜欢加布里埃尔的答案，但我改变了它的灵活性。您可以将bar-length发送到函数，并获得您想要的任何长度的进度条。进度条的长度不能为零或负。同样，你也可以像Gabriel answer一样使用这个函数(请看例子#2)。

import sys
import time

def ProgressBar(Total, Progress, BarLength=20, ProgressIcon="#", BarIcon="-"):
    try:
        # You can't have a progress bar with zero or negative length.
        if BarLength <1:
            BarLength = 20
        # Use status variable for going to the next line after progress completion.
        Status = ""
        # Calcuting progress between 0 and 1 for percentage.
        Progress = float(Progress) / float(Total)
        # Doing this conditions at final progressing.
        if Progress >= 1.:
            Progress = 1
            Status = "\r\n"    # Going to the next line
        # Calculating how many places should be filled
        Block = int(round(BarLength * Progress))
        # Show this
        Bar = "[{}] {:.0f}% {}".format(ProgressIcon * Block + BarIcon * (BarLength - Block), round(Progress * 100, 0), Status)
        return Bar
    except:
        return "ERROR"

def ShowBar(Bar):
    sys.stdout.write(Bar)
    sys.stdout.flush()

if __name__ == '__main__':
    print("This is a simple progress bar.\n")

    # Example #1:
    print('Example #1')
    Runs = 10
    for i in range(Runs + 1):
        progressBar = "\rProgress: " + ProgressBar(10, i, Runs)
        ShowBar(progressBar)
        time.sleep(1)

    # Example #2:
    print('\nExample #2')
    Runs = 10
    for i in range(Runs + 1):
        progressBar = "\rProgress: " + ProgressBar(10, i, 20, '|', '.')
        ShowBar(progressBar)
        time.sleep(1)

    print('\nDone.')

# Example #2:
Runs = 10
for i in range(Runs + 1):
    ProgressBar(10, i)
    time.sleep(1)

结果:

这是一个简单的进度条。 示例# 1 进度:[###-------]30% 例# 2 进步 : [||||||||||||........) 60% 完成了。

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)

from IPython.display import clear_output
progress_bar=u"\u001b[7m Loading: "
for i in range(100):
    clear_output(wait=True)
    progress_bar+=u"\u001b[7m "
    print(progress_bar+str(i+1)+"%")
    time.sleep(0.03) #you can change the speed

输出