如何打印多行,让后续行独立替换每一行 (Python)

How can I print multiple lines, having subsequent lines replacing each of them independently (Python)

我正在运行使用多处理模块中的 Pool 对象并行工作的程序。

我想做的是 运行 一个函数并行 n 次,每次都有一个单独的加载百分比,我希望它更新每个函数的百分比而不替换其他百分比... 示例:

f(x):
    while x < 0:
        print 'For x = {0}, {1}% completed...\r'.format(x, percentage),

而且我会 运行 并行多次执行该函数。

我想要实现的效果如下,f(10000000), f(15000000), f(7000000):

For x = 10000000, 43% completed
For x = 15000000, 31% completed
For x = 7000000, 77% completed

并且百分比将在各自的行中更新而不替换 x 的其他值,在此函数 f 中将同时 运行ning 三次。

我尝试使用回车 return '\r' 但这会替换每一行并且只会造成混乱。

感谢您抽出宝贵时间阅读本文post!希望你能告诉我。

我正在使用 Python 2.7,但如果它只能通过 Python 3 实现,我愿意接受建议。

诅咒

正如@Keozon 在评论中提到的,实现这一目标的一种方法是使用 curses 库。

python 网站上有一个 good guide for curses

ANSI 转义码

或者,您可以尝试使用 ANSI escape codes 来移动光标。

这在 Python3 中,但它在任何版本中都可以正常工作,您只需要更改周围的打印语句(或 from __future__ import print_function)。

print('Hello')                                     
print('World')                                     
                                                   
print('3[F3[F3[K', end='') # Up, Up, Clear line
                                                   
# Cursor is at the 'H' of 'Hello'                  
                                                   
print('Hi') # Overwriting 'Hello'                  
                                                   
# Cursor is at the 'W' of 'World'                  
                                                   
print('3[E', end='') # Down                     
                                                   
# Cursor is on the blank line after 'World'        
                                                   
print('Back to the end')   

                    
              

输出:

Hi         
World          
Back to the end

编辑:

我在这里为你做了太多的工作,但是嘿,这里基本上是一个使用我上面提到的 ANSI 方法的完整解决方案:

import time
import random


class ProgressBar:
    def __init__(self, name):
        self._name = name
        self._progress = 0

    @property
    def name(self):
        return self._name

    def get_progress(self):
        """
        Randomly increment the progress bar and ensure it doesn't go
        over 100
        """
        self._progress += int(random.random()*5)
        if self._progress > 100:
            self._progress = 100

        return self._progress


class MultipleProgressBars:
    def __init__(self, progress_bars):
        self._progress_bars = progress_bars
        self._first_update = True
        self._all_finished = False

    @property
    def all_finished(self):
        """
        A boolean indicating if all progress bars are at 100
        """
        return self._all_finished

    def update(self):
        """
        Update each progress bar
        """
        # We don't want to move up and clear a line on the first run
        # so we have a flag to make sure this only happens on later
        # calls
        if not self._first_update:
            # Move up and clear the line the correct number of times
            print('3[F3[K'*len(self._progress_bars),end='', sep='')

        num_complete = 0  # Number of progress bars complete
        for progress_bar in self._progress_bars:
            name = progress_bar.name
            progress = progress_bar.get_progress()

            if progress == 100:
                num_complete += 1

            # Print out a progress bar (scaled to 40 chars wide)
            print(
                name.ljust(10),
                '[' + ('='*int(progress*0.4)).ljust(40) + ']',
                str(progress)+'%')

        if num_complete == len(self._progress_bars):
            self._all_finished = True

        self._first_update = False  # Mark the first update done


# Create a list of ProgressBars and give them relevant names
progress_bars = [
    ProgressBar('James'),
    ProgressBar('Bert'),
    ProgressBar('Alfred'),
    ProgressBar('Frank')
]

# Create a new instance of our MultipleProgressBars class
mpb = MultipleProgressBars(progress_bars)

# Keep updating them while at least one of them is still active
while not mpb.all_finished:
    mpb.update()
    time.sleep(0.2)