下面的代码是一个相当通用的解决方案，也有一个时间消耗和剩余时间估计。你可以使用任何可迭代对象。进度条的大小固定为25个字符，但可以使用完整、半块和四分之一块字符以1%的速度显示更新。输出如下所示:

 18% |████▌                    | \ [0:00:01, 0:00:06]

代码示例:

import sys, time
from numpy import linspace

def ProgressBar(iterObj):
  def SecToStr(sec):
    m, s = divmod(sec, 60)
    h, m = divmod(m, 60)
    return u'%d:%02d:%02d'%(h, m, s)
  L = len(iterObj)
  steps = {int(x):y for x,y in zip(linspace(0, L, min(100,L), endpoint=False),
                                   linspace(0, 100, min(100,L), endpoint=False))}
  qSteps = ['', u'\u258E', u'\u258C', u'\u258A'] # quarter and half block chars
  startT = time.time()
  timeStr = '   [0:00:00, -:--:--]'
  activity = [' -',' \\',' |',' /']
  for nn,item in enumerate(iterObj):
    if nn in steps:
      done = u'\u2588'*int(steps[nn]/4.0)+qSteps[int(steps[nn]%4)]
      todo = ' '*(25-len(done))
      barStr = u'%4d%% |%s%s|'%(steps[nn], done, todo)
    if nn>0:
      endT = time.time()
      timeStr = ' [%s, %s]'%(SecToStr(endT-startT),
                             SecToStr((endT-startT)*(L/float(nn)-1)))
    sys.stdout.write('\r'+barStr+activity[nn%4]+timeStr); sys.stdout.flush()
    yield item
  barStr = u'%4d%% |%s|'%(100, u'\u2588'*25)
  timeStr = '   [%s, 0:00:00]\n'%(SecToStr(time.time()-startT))
  sys.stdout.write('\r'+barStr+timeStr); sys.stdout.flush()

# Example
s = ''
for c in ProgressBar(list('Disassemble and reassemble this string')):
  time.sleep(0.2)
  s += c
print(s)

欢迎提出改进建议或其他意见。干杯!

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。

PIP安装progressbar2

import os
import time
import progressbar 

os.environ['PYCHARM_HOSTED'] = '1' # https://github.com/WoLpH/python-progressbar/issues/237

class COLOR: # https://stackoverflow.com/a/287944/11465149
    YELLOW    = '\033[93m'
    GREEN     = '\033[92m'
    RED       = '\033[91m'
    BOLD      = '\033[1m'
    ENDC      = '\033[0m'

widgets=[
    'FILE.JSON ',
    COLOR.YELLOW          , progressbar.Percentage()                        , COLOR.ENDC,
    COLOR.RED + COLOR.BOLD, progressbar.Bar(left=' ', marker='━', right=' '), COLOR.ENDC,
    COLOR.YELLOW          , progressbar.Timer()                             , COLOR.ENDC
]

for i in progressbar.progressbar(range(100), widgets=widgets):
    time.sleep(0.01)
    if i == 99:
        widgets[4] = COLOR.GREEN

使用enumerate(…progressbar(max_value=…)+ this，以防你想使用它作为下载进度条

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()

我想我有点晚了，但这应该适用于使用当前版本的python 3的人，因为这使用了“f-strings”，正如python 3.6 PEP 498中介绍的那样:

Code

from numpy import interp

class Progress:
    def __init__(self, value, end, title='Downloading',buffer=20):
        self.title = title
        #when calling in a for loop it doesn't include the last number
        self.end = end -1
        self.buffer = buffer
        self.value = value
        self.progress()

    def progress(self):
        maped = int(interp(self.value, [0, self.end], [0, self.buffer]))
        print(f'{self.title}: [{"#"*maped}{"-"*(self.buffer - maped)}]{self.value}/{self.end} {((self.value/self.end)*100):.2f}%', end='\r')

例子

#some loop that does perfroms a task
for x in range(21)  #set to 21 to include until 20
    Progress(x, 21)

输出

Downloading: [########------------] 8/20 40.00%

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