jelde015的一个更普通的答案(当然是他的功劳)

手动更新加载条将是:

import sys
from math import *


def loadingBar(i, N, size):
    percent = float(i) / float(N)
    sys.stdout.write("\r"
                     + str(int(i)).rjust(3, '0')
                     +"/"
                     +str(int(N)).rjust(3, '0')
                     + ' ['
                     + '='*ceil(percent*size)
                     + ' '*floor((1-percent)*size)
                     + ']')

称之为:

loadingBar(7, 220, 40)

将结果:

007/220 [=                                       ]  

只要你想用当前的I值调用它。

将大小设置为条形图应有的字符数

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。

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)

这个进度条显示了每完成2%的点和每完成10%的数字。

import sys

def ProgressBar (num, total, nextPercent, nextPoint):
    num = float (num)
    total = float (total) - 1
    if not nextPoint:
        nextPoint = 0.0
    if not nextPercent:
        nextPoint += 2.0
        sys.stdout.write ("[0%")
        nextPercent = 10
    elif num == total:
        sys.stdout.write ("100%]\n")
        nextPercent += 10
    elif not nextPoint:
        nextPoint = 0.0
    elif num / total * 100 >= nextPercent:
        sys.stdout.write (str(int (nextPercent)) + "%")
        nextPercent += 10
    elif num / total * 100 >= nextPoint:
        sys.stdout.write (":")
        nextPoint += 2
    return (nextPercent, nextPoint)

nextPercent, nextPoint = 0, 0
total = 1000

for num in range (total):
    nextPercent, nextPoint = ProgressBar (num, total, nextPercent, nextPoint)

结果:

>>> 
[0%::::10%:::::20%:::::30%:::::40%:::::50%:::::60%:::::70%:::::80%:::::90%:::::100%]
>>> 

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")