Python搜索字符串以查找列表中任何项目的第一次出现

Question

I have to parse a few thousand txt documents using python, but right now I'm getting the code working for just one.

I am trying to find the first time any month (January, February, March, etc) appears in the document, and return the position of that first month. Every document has at least one month in it, but some have many months.

This works currently, but seems very cumbersome:

mytext = open('2.txt','r')
mytext = mytext.read()

January = mytext.find("January")
February = mytext.find("February")
March = mytext.find("March")
April = mytext.find("April")
May = mytext.find("May")
June = mytext.find("June")
July = mytext.find("July")
August = mytext.find("August")
September = mytext.find("September")
October = mytext.find("October")
November = mytext.find("November")
December = mytext.find("December")

monthpos = [January, February, March, April, May, June, July, August, September, October, November, December]
monthpos = [x for x in monthpos if x != -1]
print min(monthpos)
 # returns the first match as a number

I would like to combine something like any() and find() to get the job done, but there doesn't seem like a better way to do this. I found this question but it isn't so clear, so it didn't help that much. While I know this is wrong and does not work for many reasons, here is what I want to do:

mytext = open('text.txt','r')
mytext = mytext.read()
months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]
print mytext.find(months) #where this would find the first time any month is matched
1945 # return the location in the string where the first month is found

Thanks in advance.

Answer 1

I think this would do what you want:

months = ["January", "February", "March", "April", 
          "May", "June", "July", "August", 
          "September", "October", "November", "December"]
indices = [s.find(month) for month in months]
first = min(index for index in indices if index > -1)

First, we get the first appearance of each month (or -1 if not present), then we get the minimum of the indices, except where it's -1 . This will throw a ValueError if none are found, which may or may not be what you want.

---

As Two-Bit Alchemist has commented, you could short-cut for efficiency:

months = ["January", "February", "March", "April", 
          "May", "June", "July", "August", 
          "September", "October", "November", "December"]
first = None
for month in sorted(months, key=len):
    i = s[:first].find(month) # only search first part of string
    if i != -1:
        if i < first or first is None:
            first = i
        if i < len(month): # not enough room for any remaining months
            break

Answer 2

I would use re for conceptual simplicity. It's also easy to extend the code to do something more complex if you need to later.

import re
mytext = open('text.txt','r')
mytext = mytext.read()
months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]
months_match = re.search("|".join(months), mytext)
print match_obj.start()

http://docs.python.org/2/library/re.html