如何在python中打印列表理解的进度? [英] How to print the progress of a list comprehension in python?

问题描述

在我的方法中，我必须返回列表中的列表.我想要一个列表理解功能，因为它的性能很高，因为创建列表大约需要5分钟.

In my method i have to return a list within a list. I would like to have a list comprehension, because of the performance since the list takes about 5 minutes to create.

[[token.text for token in document] for document in doc_collection]

是否可以打印进度，当前创建过程位于哪个文档中?像这样的东西:

Is there a possibility to print out the progress, in which document the create-process currently are? Something like that:

[[token.text for token in document] 
  and print(progress) for progress, document in enumerate(doc_collection)]

感谢您的帮助！

推荐答案

1:使用辅助功能

def report(index):
    if index % 1000 == 0:
        print(index)

def process(token, index, report=None):
    if report:
        report(index) 
    return token['text']

l1 = [{'text': k} for k in range(5000)]

l2 = [process(token, i, report) for i, token in enumerate(l1)]

2:使用and和or语句

2: Use and and or statements

def process(token):
    return token['text']

l1 = [{'text': k} for k in range(5000)]
l2 = [(i % 1000 == 0 and print(i)) or process(token) for i, token in enumerate(l1)]

3:同时使用

def process(token):
    return token['text']

def report(i):
    i % 1000 == 0 and print(i)

l1 = [{'text': k} for k in range(5000)]
l2 = [report(i) or process(token) for i, token in enumerate(l1)]

---

所有3种方法均可打印:

---

All 3 methods print:

0
1000
2000
3000
4000

---

2的工作方式

• i % 1000 == 0 and print(i):and仅在第二条语句为True时检查第二条语句，因此仅在i % 1000 == 0
时打印
• or process(token):or始终检查两个语句，但返回评估为True的第一个语句.
  • 如果为i % 1000 != 0，则第一个语句为False，并且将process(token)添加到列表中.
  • 否则，第一个语句为None(因为print返回None)，同样，or语句将process(token)添加到列表中
  • i % 1000 == 0 and print(i): and only checks the second statement if the first one is True so only prints when i % 1000 == 0
  • or process(token): or always checks both statements, but returns the first one which evals to True.
    • If i % 1000 != 0 then the first statement is False and process(token) is added to the list.
    • Else, then the first statement is None (because print returns None) and likewise, the or statement adds process(token) to the list

    3种工作原理

    与2相似，因为report(i)没有任何return内容，因此它等效于None，并且or将process(token)添加到列表中

    Similarly as 2, because report(i) does not return anything, it evals to None and or adds process(token) to the list

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