问Python列表理解个人挑战
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Stack Overflow用户

提问于 2014-06-29 17:58:11

回答 3查看 947关注 0票数 0

给定一个文本文件"words.txt"，使用列表理解来读取文件中的所有单词，并找到至少包含两个元音的所有单词。

所以，我有一个文本文件：

The quick brown fox jumps over the lazy dog

而且，获得所有单词和所有具有两个或多个元音的单词的最佳尝试是：

#This could be hardcoded in, but for the sake of simplicity (as simple as simplicity gets)
vowels = ["a","e","i","o","u"]
filename = "words.txt"
words = [word for word in open(filename, "r").read().split()]
multivowels = [each for each in open(filename, "r").read().split() if sum(letter in vowels for letter in each) >= 2]

输出应模仿：

All words in the file:  ['The', 'quick', 'brown', 'fox', 'jumps', 'over', 'the', 'lazy', 'dog']
The words in the file that contain 2 or more vowels: ['quick', 'over']

我试图把它写成一行，只是打印“单词”和“多元音”以及“文件中的所有单词”的列表理解部分。等。

有谁愿意接受把这两种理解结合在一起的挑战呢？我和我的队友都很困惑，但很乐意向我们的教授展示！

同样，我的最后一行代码是：

print "All words in the file: " + str([word for word in open(filename, "r").read().split()]) + "\nAll words with more than 2 vowels: " + str([each for each in open(filename, "r").read().split() if sum(letter in vowels for letter in each) >= 2])

编辑:我试图获取文件中的所有单词，以及所有带有两个或多个元音的单词。

vowels = ["a", "e", "i", "o", "u"]
filename = "words.txt"
print [(word, each) for word in open(filename, "r").read().split() if sum([1 for each in word if each in vowels]) >= 2]

python

file

list-comprehension

word

回答 3

Stack Overflow用户

发布于 2014-06-29 19:22:47

这里有一些需要处理的角落案例，但是如果您假设一个简单的文本文件：

import re
vowels = "a","e","i","o","u"

answer = [[word for word in re.sub("[^\w]", " ",  sentence).split() if (sum(1 for letter in word if letter in vowels)>=2)] for sentence in open(filename,"r").readlines()]

票数 1

Stack Overflow用户

发布于 2014-06-30 07:04:20

我在一个更大的数据集中运行了一个稍微不同的版本，并发现对str的重复调用开始增加。

vowels = ['a', 'e', 'i', 'o', 'u']
#filename = 'vowelcount.txt'
filename = 'largetextfile.txt'
print "All words in the file: ", [w for w in open(filename).read().split()], "\n", "All words with more than 2 vowels: ", [w for w in open(filename).read().split() if sum(1 for l in w if l in vowels) > 1]

用cProfile调用此版本显示了一个小小的改进：

python -m cProfile vowelcount1.py

   7839 function calls in 0.023 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.005    0.005    0.023    0.023 vowelcount1.py:1(<module>)
     6045    0.009    0.000    0.009    0.000 vowelcount1.py:4(<genexpr>)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
        2    0.000    0.000    0.000    0.000 {method 'read' of 'file' objects}
        2    0.000    0.000    0.000    0.000 {method 'split' of 'str' objects}
        2    0.000    0.000    0.000    0.000 {open}
     1786    0.009    0.000    0.018    0.000 {sum}

您输入的代码几乎是函数调用数的两倍：

#filename = 'vowelcount.txt'
filename = 'largetextfile.txt'
vowels = ['a', 'e', 'i', 'o', 'u']
print "All words in the file: " + str([word for word in open(filename, "r").read().split()]) + "\nAll words with more than 2 vowels: " + str([each for each in open(filename, "r").read().split() if sum(letter in vowels for letter in each) >= 2])

python -m cProfile vowelcount2.py

   14568 function calls in 0.036 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.004    0.004    0.036    0.036 vowelcount2.py:2(<module>)
    12774    0.016    0.000    0.016    0.000 vowelcount2.py:4(<genexpr>)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
        2    0.000    0.000    0.000    0.000 {method 'read' of 'file' objects}
        2    0.000    0.000    0.000    0.000 {method 'split' of 'str' objects}
        2    0.000    0.000    0.000    0.000 {open}
     1786    0.016    0.000    0.032    0.000 {sum}