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")

没有外部包。一段现成的代码。

您可以自定义进度条符号“#”，进度条大小，文本前缀等。

Python 3.3 +

import sys
def progressbar(it, prefix="", size=60, out=sys.stdout): # Python3.3+
    count = len(it)
    def show(j):
        x = int(size*j/count)
        print("{}[{}{}] {}/{}".format(prefix, "#"*x, "."*(size-x), j, count), 
                end='\r', file=out, flush=True)
    show(0)
    for i, item in enumerate(it):
        yield item
        show(i+1)
    print("\n", flush=True, file=out)

用法:

import time    
for i in progressbar(range(15), "Computing: ", 40):
    time.sleep(0.1) # any code you need

要填满整个字符空间，请使用unicode u"█"字符替换"#"。使用for i in progressbar(range(100)):…你会得到:

不需要第二个线程。上面的一些解决方案/包需要。 适用于任何可迭代对象，它指的是任何可以使用len()的对象。一个列表，一个字典，比如[' A '， 'b'， 'c'…' g '] 使用生成器的工作只需要用list()来包装它。例如，对于i in progressbar(list(your_generator)， "Computing: "， 40):除非工作在生成器中完成。在这种情况下，您需要另一种解决方案(如tqdm)。

您还可以通过将out更改为sys来更改输出。例如Stderr。

Python 3.6+ (f-string)

def progressbar(it, prefix="", size=60, out=sys.stdout): # Python3.6+
    count = len(it)
    def show(j):
        x = int(size*j/count)
        print(f"{prefix}[{u'█'*x}{('.'*(size-x))}] {j}/{count}", end='\r', file=out, flush=True)
    show(0)
    for i, item in enumerate(it):
        yield item
        show(i+1)
    print("\n", flush=True, file=out)

Python 2 (old-code)

import sys
def progressbar(it, prefix="", size=60, out=sys.stdout):
    count = len(it)
    def show(j):
        x = int(size*j/count)
        out.write("%s[%s%s] %i/%i\r" % (prefix, u"#"*x, "."*(size-x), j, count))
        out.flush()        
    show(0)
    for i, item in enumerate(it):
        yield item
        show(i+1)
    out.write("\n")
    out.flush()

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。

使用os_sys lib:

我在许多类型的条形图中使用它，例如:

from os_sys.progress import bar as Bar
bar = Bar('progresing: ', max=20)
for i in range(20):
    #do somthing
    bar.next()
bar.finish()

你的输出将是:

procesing:  |######                          | 2/10

请在os_sys的描述中阅读更多信息

#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

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