Question

我正在编写一个程序，该文件从50,000个单词的文件中读取，并且需要获得其中没有字母“e”的单词的百分比。我可以让程序打印出没有e的所有单词，但我想将它们附加到列表中，以便我可以得到列表中元素的总和。我现在拥有的每次运行时都会得到0的结果。它还产生正确的总线数。对不起，我不是python中最好的。

f=open("hardwords.txt")

def has_no_e(f):
    words = []
    sum_words= len(words)
    total = sum(1 for s in f)
    print total
    print sum_words
    letter = 'e'
    for line in f:
        for l in letter:
            if l in line:
                break
        else:
            words.append(line)

has_no_e(f)

Answer 1

你不需要收集单词，只计算它们。

未测试：

total = 0
without_e = 0
with open("hardwords.txt") as f:
    for line in f:
        total = total + 1
        if not 'e' in line:
            without_e = without_e + 1

percentage = float(without_e) / float(total)

Answer 2

这个怎么样：

def has_no_e():
    with open(path, "r") as f:
        words = [word.strip() for line in f.readlines() for word in line.strip().split(',')]
        words_without_e = [word for word in words if 'e' not in word]
        print len(words), words
        print len(words_without_e), words_without_e

has_no_e()

现在你只需要计算百分比

Answer 3

就是这样：

def has_no_e(path):
    total_words = 0
    words_without_e = 0
    with open(path, "r") as f:
        for line in f:
            words = line.lower().split()
            total_words += len(words)
            words_without_e += sum("e" not in w for w in words)

    return (float(words_without_e)/total_words)*100

Answer 4

这是一种可行的方法：

with open('G:\Tmp\demo.txt', 'r') as f:
    total = 0
    count = 0
    for line in f:
        words = line.split()
        total = total + len(words)
        count = count + len([w for w in words if w.find('e') > 0])

print 'Total word:{0}, counted:{1}'.format(total, count)

将特定单词附加到python

4 个答案: