如何从几个文本文件中获取元组列表? [英] How to get a list of tuples from several text files?

问题描述

我想访问46个子目录中的.txt文件，并提取每个文件文本中的0和1.到目前为止，我已经编写了以下代码:

I want to access .txt files in 46 subdirectories and extract the number of 0s and 1s in the text of each file. So far I've written this code:

from pathlib import Path

def count_0s(paths):
  for p in paths:
    list_zeros = []
    list_ones = []
    for line in p.read_text().splitlines():
      zeros = 0
      zeros += line.count('0')
      ones = 0
      ones += line.count('1')
    list_zeros.append(zeros)
    list_ones.append(ones)    
  return list_zeros, list_ones

path = "/content/drive/MyDrive/data/classes/"
paths = Path(path).glob("*/marked*.txt")
n_zeros=count_0s(paths)
n_zeros

我想以2个列表的形式返回函数(一个带有0的数目，另一个带有1的数目)以在Pandas数据框中使用.对不起，如果重复的问题.

I want to get the function return in the form of 2 lists (one with the number of 0s and the other with the number of 1s) to use in a Pandas dataframe. Sorry if the questions are duplicated.

推荐答案

函数中存在一些错误:

• 您添加了一些不必要的方括号( splitlines()已返回列表)
• 您不需要遍历字符，而是遍历行

这是一个更正的功能:

def count_0s(paths):
  zeros_list = []
  ones_list = []
  for p in paths:
    zeros = 0
    ones = 0
    for line in p.read_text().splitlines():
        for c in line:
            if c == '0':
                zeros += 1
            else:
                ones += 1
    zeros_list.append(zeros)
    ones_list.append(ones)
  return zeros_list, ones_list

请注意，这可能是计算0和1的效率很低的方法.例如，仅使用 line.count('0')而不是for循环可以使速度提高一倍之10.

Be aware that this is probably a very inefficient way of counting 0 and 1. For example just using line.count('0') instead of a for loop can increase the speed by a factor of 10.

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