python TypeError：str对象不是迭代器

Question

我正在编写一个在文本文件中搜索值并打印上一行的代码； 但我遇到了两个问题，首先我没有找到打印上一行的方法，当我编写代码打印下一个值时出现错误。

代码：-

with open('Perfix/prefix.txt') as f:
    lines = f.readlines()
    for line in lines:
        if "100" in line:
            print (next(line))

假设场景：-

代码应该在 prefix.txt 文件中搜索以找到值 100 的行，然后它应该打印上一行。就这样！！

错误：-

Traceback (most recent call last):
  File "1.py", line 7, in <module>
    print next(line)
TypeError: str object is not an iterator

什么是错误？另外，您能否建议如何打印“上一行”而不是下一行？

示例：-

文件前缀.txt

00
122
141
1525
1162
1547
100
125
15321
1100
1513
142100

输出：-

1547

Answer 1

查找前一行的最简单方法是在每次不匹配时存储该行：

with open('Perfix/prefix.txt') as f:
    previous = None
    for line in f:
        line = line.strip()
        if line == '100':
            print previous
        previous = line

请注意，您不需要先将所有行读入内存；只需遍历文件对象。

next() 函数根本不应该在这里使用；它需要一个迭代器，并将其推进到下一个项，而不是将其倒回到上一个条目。

Answer 2

错误消息很清楚 - line 是 string，而不是迭代器，因此 next(line) 没有意义。考虑：

next('hello')

应该发生什么？

---

为了解决您的问题，我建议如下：

for index, line in enumerate(lines): # get the index for each line as we go
    if index and line.strip() == '100': # if this isn't the first line & is just '100'
        print lines[index-1] # print the previous line
        break # optional - stop iterating if you don't want to find further lines

或成对遍历行；参见例如Iterate a list as pair (current, next) in Python.

另请注意，您可以在文件关闭后执行此操作：

 with open(...) as f:
     lines = f.readlines()
 for index, line in enumerate(lines):
     ...

Answer 3

通过re 模块。

#!/usr/bin/python
import re
with open('file', 'r') as f:
    lines = f.read()
    print re.findall(r'(?m)^(.*)\n100$', lines)[0]

输出：

1547

说明：

(?m)                     set flags for this block (with ^ and $
                         matching start and end of line) (case-
                         sensitive) (with . not matching \n)
                         (matching whitespace and # normally)
^                        the beginning of a "line"
(                        group and capture to \1:
  .*                       any character except \n (0 or more
                           times)
)                        end of \1
\n                       '\n' (newline)
100                      '100'
$                        before an optional \n, and the end of a
                         "line"