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")

这是创建进度条的简单方法

import time,sys
toolbar_width = 50
# setting up toolbar [-------------------------------------]
sys.stdout.write("[%s]"%(("-")*toolbar_width))
sys.stdout.flush()
# each hash represents 2 % of the progress
for i in range(toolbar_width):
    sys.stdout.write("\r") # return to start of line
    sys.stdout.flush()
    sys.stdout.write("[")#Overwrite over the existing text from the start 
    sys.stdout.write("#"*(i+1))# number of # denotes the progress completed 
    sys.stdout.flush()
    time.sleep(0.1)

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")

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。

这在Python3中非常简单:

   import time
   import math

    def show_progress_bar(bar_length, completed, total):
        bar_length_unit_value = (total / bar_length)
        completed_bar_part = math.ceil(completed / bar_length_unit_value)
        progress = "*" * completed_bar_part
        remaining = " " * (bar_length - completed_bar_part)
        percent_done = "%.2f" % ((completed / total) * 100)
        print(f'[{progress}{remaining}] {percent_done}%', end='\r')

    bar_length = 30
    total = 100
    for i in range(0, total + 1):
        show_progress_bar(bar_length, i, total)
        time.sleep(0.1)

    print('\n')