Question

运行时，光标在左侧的命令行上闪烁，没有任何反应。谁能告诉我为什么它没有做到预期的呢？

感谢您的期待。

代码如下：

import sys


i=3   
try:
    text = open(sys.argv[1], 'r')

except IOError:
  print 'Cannot open file %s for reading' % filename

  sys.exit(0)



char = text.read (1)

#navigate to ith sentence, i.e. go to the first letter of the ith sentence

j = 0
for j in range (0, i-1):
    char = text.read (1)
    if char == '.':
        j = j+1

char = text.read(2) #Assuming there is a spce after full-stop/!/?

#count the number of characters in the present sentence

chars = 0

while char != '.' or '!' or '?':
    char = text.read (1)
    chars = chars + 1


print chars

Answer 1

我看到一个问题：

while char != '.' or '!' or '?':

应该是：

while char != '.' and char != '!' and char != '?':

...或至少在逻辑上等同的东西，但更清洁，更美味。在前一种情况下，Python将尝试评估char != '.'，然后评估'!'，并评估'?'，两者都被视为true（因为它们不等于零）。所以循环会永远持续下去！

至于：

j = 0
for j in range (0, i-1):
    char = text.read (1)
    if char == '.':
        j = j+1

我认为这不符合你的意图。假设文本文件就像'这是一个句子。这是另一句话。'，该循环将char设置为'i'，j设置为1.您可能打算使用while循环，如上所述，而不是单个if语句。

Answer 2

当你到达文件的末尾时，read（1）每次都会返回一个空字符串

    Help on built-in function read:

    read(...)
        read([size]) -> read at most size bytes, returned as a string.

        If the size argument is negative or omitted, read until EOF is reached.
        Notice that when in non-blocking mode, less data than what was requested
        may be returned, even if no size parameter was given.

使用Python查找给定文本文件的第i个句子中的字符数

2 个答案: