使用正則表達式提取數據：Python

Question

該問題的基本概述是讀取文件，使用re.findall()查找整數，查找[0-9]+的正則表達式，然後將提取的字符串轉換爲整數併合計整數。

我在追加列表中遇到麻煩。從我的下面的代碼，它只是追加行的第一個（0）索引。請幫幫我。謝謝。

import re 
hand = open ('a.txt') 
lst = list() 
for line in hand: 
    line = line.rstrip() 
    stuff = re.findall('[0-9]+', line) 
    if len(stuff)!= 1 : continue 
    num = int (stuff[0]) 
    lst.append(num) 
print sum(lst) 

Answer 1

很好，謝謝你包括整個txt文件！你的主要問題是if len(stuff)...這條線跳過，如果stuff沒有東西和它有2,3等等。您只保留stuff長度爲1的列表。我在代碼中添加了註釋，但如果有不明之處，請提出任何問題。

import re 
hand = open ('a.txt') 
str_num_lst = list() 
for line in hand: 
    line = line.rstrip() 
    stuff = re.findall('[0-9]+', line) 
    #If we didn't find anything on this line then continue 
    if len(stuff) == 0: continue 
    #if len(stuff)!= 1: continue #<-- This line was wrong as it skip lists with more than 1 element 

    #If we did find something, stuff will be a list of string: 
    #(i.e. stuff = ['9607', '4292', '4498'] or stuff = ['4563']) 
    #For now lets just add this list onto our str_num_list 
    #without worrying about converting to int. 
    #We use '+=' instead of 'append' since both stuff and str_num_lst are lists 
    str_num_lst += stuff 

#Print out the str_num_list to check if everything's ok 
print str_num_lst 

#Get an overall sum by looping over the string numbers in the str_num_lst 
#Can convert to int inside the loop 
overall_sum = 0 
for str_num in str_num_lst: 
    overall_sum += int(str_num) 

#Print sum 
print 'Overall sum is:' 
print overall_sum 

編輯：

你是對的，在整個文件中讀取爲一條線是一個很好的解決方案，這是不難做到。檢查出this post。這是代碼的樣子。

import re 

hand = open('a.txt') 
all_lines = hand.read() #Reads in all lines as one long string 
all_str_nums_as_one_line = re.findall('[0-9]+',all_lines) 
hand.close() #<-- can close the file now since we've read it in 

#Go through all the matches to get a total 
tot = 0 
for str_num in all_str_nums_as_one_line: 
    tot += int(str_num) 

print 'Overall sum is:',tot