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()

试试https://pypi.python.org/pypi/progress上的progress。

from progress.bar import Bar

bar = Bar('Processing', max=20)
for i in range(20):
    # Do some work
    bar.next()
bar.finish()

结果将是如下所示的条:

Processing |#############                   | 42/100

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")

#doesnt affect actual execution
#based on events and consumption in background
#may be that actual process completes a bit earlier than progress shows 99%
#make an instance with number of elements in a loop
#in each iteration call the method current_progress

import time
from math import ceil
import os
import sys
from threading import Thread
class progress_bar(object):
 def __init__(self,total_elements,length_bar=25):
  self.length_bar=length_bar
  self.total_elements=total_elements
  self.singleweight=(float(1)/float(total_elements))*100
  self.done=0
  self.qt=[0]
  self.call_count=0
  t=Thread(target=self.display_progress)
  t.start()
 def current_progress(self):
  self.done+=1
  self.qt=[self.done]+self.qt
 def display_progress(self):
  while True:
   try:
    done=self.qt.pop()
   except:
    continue
   else:
    self.call_count+=1
    self.progress=self.singleweight*done
    fill=ceil(self.progress)
    bar=int((fill*self.length_bar)/100)*"|"
    bar="["+bar+str(fill)+"%"
    barp=bar
    for i in range(0,self.length_bar+3-(len(bar))):
     barp=barp+"_"
    barp=barp+"]"
    if self.progress <= 100:
     os.system("clear")
     print("Progress:",barp, sep=' ', end='\n', file=sys.stdout, flush=True)
    if self.call_count == self.total_elements:
     break
  else:
   pass

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!