Python 搜索文件并找到精确匹配和打印行？
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Search File And Find Exact Match And Print Line?
提问by Robots
I searched around but i couldn't find any post to help me fix this problem, I found similar but i couldn't find any thing addressing this alone anyway.
我四处搜索，但找不到任何帖子来帮助我解决这个问题，我发现了类似的问题，但无论如何我都找不到任何单独解决此问题的方法。
Here's the the problem I have, I'm trying to have a python script search a text file, the text file has numbers in a list and every number corresponds to a line of text and if the raw_input match's the exact number in the text file it prints that whole line of text. so far It prints any line containing the the number.
这是我遇到的问题，我正在尝试让 python 脚本搜索文本文件，文本文件在列表中有数字，每个数字对应一行文本，如果 raw_input 匹配文本文件中的确切数字它打印整行文本。到目前为止，它打印包含数字的任何行。
Example of the problem, User types 20then the output is every thing containing a 2and a 0, so i get 220 foo200 baretc. How can i fix this so it just find "20"
问题的例子，用户输入20然后输出是包含 a2和 a 的所有东西0，所以我得到220 foo200 bar等等。我该如何解决这个问题，所以它只找到“20”
here is the code i have
这是我的代码
num = raw_input ("Type Number : ")
search = open("file.txt")
for line in search:
if num in line:
print line
Thanks.
谢谢。
回答by unutbu
To check for an exactmatch you would use num == line. But linehas an end-of-line character \nor \r\nwhich will not be in numsince raw_inputstrips the trailing newline. So it may be convenient to remove all whitespace at the end of linewith
要检查完全匹配，您可以使用num == line. 但是line有一个行尾字符，\n或者因为去掉了尾随的换行符\r\n而不会出现。因此删除with末尾的所有空格可能会很方便numraw_inputline
line = line.rstrip()
with open("file.txt") as search:
for line in search:
line = line.rstrip() # remove '\n' at end of line
if num == line:
print(line )
回答by naeg
The check has to be like this:
支票必须是这样的：
if num == line.split()[0]:
If file.txt has a layout like this:
如果 file.txt 有这样的布局：
1 foo
20 bar
30 20
We split up "1 foo"into ['1', 'foo']and just use the first item, which is the number.
我们分裂"1 foo"成['1', 'foo']，只是使用的第一个项目，这是多少。
回答by Oni1
It's very easy:
这很容易：
numb = raw_input('Input Line: ')
fiIn = open('file.txt').readlines()
for lines in fiIn:
if numb == lines[0]:
print lines
回答by lenik
you should use regular expressions to find all you need:
你应该使用正则表达式来找到你需要的一切：
import re
p = re.compile(r'(\d+)') # a pattern for a number
for line in file :
if num in p.findall(line) :
print line
regular expression will return you all numbers in a line as a list, for example:
正则表达式会将一行中的所有数字作为列表返回，例如：
>>> re.compile(r'(\d+)').findall('123kh234hi56h9234hj29kjh290')
['123', '234', '56', '9234', '29', '290']
so you don't match '200' or '220' for '20'.
所以你不匹配 '200' 或 '220' 为 '20'。
回答by Jetlef
num = raw_input ("Type Number : ")
search = open("file.txt","r")
for line in search.readlines():
for digit in num:
# Check if any of the digits provided by the user are in the line.
if digit in line:
print line
continue
回答by The Aelfinn
Build lists of matched lines - several flavors:
构建匹配行的列表 - 几种风格：
def lines_that_equal(line_to_match, fp):
return [line for line in fp if line == line_to_match]
def lines_that_contain(string, fp):
return [line for line in fp if string in line]
def lines_that_start_with(string, fp):
return [line for line in fp if line.startswith(string)]
def lines_that_end_with(string, fp):
return [line for line in fp if line.endswith(string)]
Build generator of matched lines (memory efficient):
构建匹配行的生成器（内存效率高）：
def generate_lines_that_equal(string, fp):
for line in fp:
if line == string:
yield line
Print all matching lines (find all matches first, then print them):
打印所有匹配的行（首先找到所有匹配，然后打印它们）：
with open("file.txt", "r") as fp:
for line in lines_that_equal("my_string", fp):
print line
Print all matching lines (print them lazily, as we find them)
打印所有匹配的行（在我们找到它们时懒惰地打印它们）
with open("file.txt", "r") as fp:
for line in generate_lines_that_equal("my_string", fp):
print line
Generators (produced by yield) are your friends, especially with large files that don't fit into memory.
生成器（由 yield生成）是您的朋友，尤其是对于不适合内存的大文件。
