受到许多不依赖包的答案的启发，我在这里分享我的实现。在任何循环中使用的函数都需要当前迭代数、迭代总数和初始时间。

import time    
def simple_progress_bar(i: int, n: int, init_time: float):
    avg_time = (time.time()-init_time)/(i+1)
    percent = ((i+1)/(n))*100
    print(
        end=f"\r|{'='*(int(percent))+'>'+'.'*int(100-int(percent))}|| " + \
        f"||Completion: {percent : 4.3f}% || \t "+ \
        f"||Time elapsed: {avg_time*(i+1):4.3f} seconds || \t " + \
        f"||Remaining time: {(avg_time*(n-(i+1))): 4.3f} seconds."
    )
    return



N = 325
t0 = time.time()
for k in range(N):
    # stuff goes here #
    time.sleep(0.0001)
    # stuff goes here #
    
    simple_progress_bar(k, N, t0)

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()

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%

from IPython.display import clear_output
progress_bar=u"\u001b[7m Loading: "
for i in range(100):
    clear_output(wait=True)
    progress_bar+=u"\u001b[7m "
    print(progress_bar+str(i+1)+"%")
    time.sleep(0.03) #you can change the speed

输出

已经有很多好的答案，但添加这个特定的基于@HandyGold75的答案，我希望它在特定的上下文中是callabe，有一个初始的msg，加上在结束时的几秒钟的时间反馈。

from time import sleep, time


class ProgressBar:
    def __init__(self, total: float, width: int = 50, msg: str = ""):
    self.total = total
    self.width = width
    self.start: float = time()
    if msg:
        print(f"{msg}")

    def progress(self, progress: float):
        percent = self.width * ((progress) / self.total)
        bar = chr(9608) * int(percent) + "-" * (self.width - int(percent))
        print(
            f"\r|{bar}| {(100/self.width)*percent:.2f}% "
            f"[{progress} of {self.total}]",
            end="\r",
        )

    def __enter__(self):
        return self.progress

    def __exit__(self, type, value, traceback):
        end: float = time()
        print(f"\nFinished after {end - self.start: .3f} seconds.")


# USAGE
total_loops = 150
with ProgressBar(total=total_loops) as progress:
    for i in range(total_loops):
        sleep(0.01)  # Do something usefull here
        progress(i + 1)

如果它是一个大循环，迭代次数固定，需要花费很多时间，你可以使用我做的这个函数。循环的每一次迭代都增加了进展。其中count是循环的当前迭代，total是循环的值，size(int)是你想要的条的大小，以10为增量，即(size 1 =10个字符，size 2 =20个字符)

import sys
def loadingBar(count,total,size):
    percent = float(count)/float(total)*100
    sys.stdout.write("\r" + str(int(count)).rjust(3,'0')+"/"+str(int(total)).rjust(3,'0') + ' [' + '='*int(percent/10)*size + ' '*(10-int(percent/10))*size + ']')

例子:

for i in range(0,100):
     loadingBar(i,100,2)
     #do some code 

输出:

i = 50
>> 050/100 [==========          ]