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%

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')

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")

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")

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

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。