Python:如何使用regex显示文本文件中的前几个数字

从我上面的评论来看:

1. 读取两个文件并将它们的行存储在列表中

2. 平展列表

3. 按字符串中的视图对列表排序

因此 :

list.txt文件:

file Marvel/GuardiansOfGalaxy 300 1
file DC/Batman 504 1
file GameOfThrones 900 0
file DC/Superman 200 1
file Marvel/CaptainAmerica 342 0

列表2.txt:

file Science/Biology 200 1
file Math/Calculus 342 0
file Psychology 324 1
file Anthropology 234 0
file Science/Chemistry 444 1

以及 :

fileOne = 'list.txt'
fileTwo = 'list2.txt'

result = []
with open (fileOne, 'r') as file1Obj, open(fileTwo, 'r') as file2Obj:
      result.append(file1Obj.readlines())
      result.append(file2Obj.readlines())

result = sum(result, [])                 # flattening the nested list
result = [i.split('\n', 1)[0] for i in result]  # removing the \n char

print(sorted(result, reverse=True, key = lambda x: int(x.split()[2]))) # sorting by the view

输出 :

[
 'file GameOfThrones 900 0', 'file DC/Batman 504 1', 'file Science/Chemistry 444 1', 
 'file Marvel/CaptainAmerica 342 0', 'file Math/Calculus 342 0', 
 'file Psychology 324 1', 'file Marvel/GuardiansOfGalaxy 300 1', 
 'file Anthropology 234 0', 'file DC/Superman 200 1', 'file Science/Biology 200 1'
]

较短版本 :

with open (fileOne, 'r') as file1Obj, open(fileTwo, 'r') as file2Obj: result = file1Obj.readlines() + file2Obj.readlines()    
print(list(i.split('\n', 1)[0] for i in sorted(result, reverse=True, key = lambda x: int(x.split()[2]))))   # sorting by the view

您可以使用以下代码:

#open the 2 files in read mode
with open('file1.txt', 'r') as f1, open('file2.txt', 'r') as f2:
  data = f1.read() + f2.read() #store the content of the two files in a string variable
  lines = data.split('\n') #split each line to generate a list
  #do the sorting in reverse mode, based on the 3rd word, in your case number of views
  print(sorted(lines[:-1], reverse=True, key=lambda x:int(x.split()[2])))

输出:

['file GameOfThrones 900 0', 'file DC/Batman 504 1', 'file Science/Chemistry 444 1', 'file Marvel/CaptainAmerica 342 0', 'file Math/Calculus 342 0', 'file Psychology 324 1', 'file Marvel/GuardiansOfGalaxy 300 1', 'file Anthropology 234 0', 'file DC/Superman 200 1', 'file Science/Biology 200 1']

提取排序条件

首先,你需要得到你想要对每一行进行排序的信息。 可以使用此正则表达式从行中提取视图和路径:

>>> import re
>>> criteria_re = re.compile(r'file (?P<path>\S*) (?P<views>\d*) \d*')
>>> m = criteria_re.match('file GameOfThrones 900 0')
>>> res = (int(m.group('views')), m.group('path'))
>>> res
(900, 'GameOfThrones')

排序

现在整个过程只需要应用到您的文件集合。因为我们不需要默认搜索,所以需要设置 key 搜索函数的参数,帮助它知道我们到底要按什么排序:

def sort_files(files):
    lines = []
    for file in records:
        for line in open(file):
            m = criteria_re.match(line)
            # maybe do some error handling here, in case the regex doesn't match
            lines.append((line, (-int(m.group('views')), m.group('path'))))
            # taking the negative view count makes the comparison later a
            # bit more simple, since we can just sort be descending order
            # for both view as well as alphabetical path order 
    # the sorting criteria were only tagging along to help with the order, so
    # we can discard them in the result
    return [line for line, criterion in sorted(lines, key=lambda x: x[1])]