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。

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。

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()

这里有一个非常简单的版本，如果你有一个循环，只是想了解迭代的进展，比如每一个点，比如说，5000次迭代。

my_list = range(0,100000)

counter = 0
for x in my_list:
    #your code here

    counter = counter + 1
    if counter % 5000 == 0:
        print(".", end="") # end="" avoids a newline, keeps dots together

print() #this makes sure whatever you print next is in a new line

My_list不是方案的一部分。使用你自己的迭代对象，不管你在循环什么。 这个版本没有提前告诉您总共有多少次迭代。