用于在一行中查找数字的正则表达式 (Python)

Question

I'm just learning about regular expressions and I need to read in a text file and find every instance of a number and find the sum of all the numbers.

import re

sum = 0
list_of_numbers = list()
working_file = open("sample.txt", 'r')
for line in working_file:
    line = line.rstrip()
    working_list = re.findall('[0-9]+', line)
    if len(working_list) != 1:
        continue
    print(working_list)
    for number in working_list:
        num = int(number)
        list_of_numbers.append(num)
for number in list_of_numbers:
    sum += number
print(sum)

I put the print(working_list) in order to try and debug it and see if all the numbers are getting found correctly and I've seen, by manually scanning the text file, that some numbers are being skipped while others are not. I'm confused as to why as I thought my regular expression guaranteed that any string with any amount of digits will be added to the list.

Here is the file .

Answer 1

你只验证只有一个数字的行，所以有两个数字的行将被跳过，因为if len(working_list) != 1: continue ，这基本上是说“如果这一行没有确切的一个数字，那么跳过”，您的意思可能类似于if len(working_list) < 1: continue

Answer 2

I would do it like:

import re

digits_re = re.compile(r'(\d+(\.\d+)?)') 
with open("sample.txt", 'r') as fh:
  numbers = [float(match[0]) for match in digits_re.findall(fh.read())]
print(sum(numbers))

or like you're doing with ints just

import re

digits_re = re.compile(r'(\d+)') 
with open("sample.txt", 'r') as fh:
  numbers = [int(match[0]) for match in digits_re.findall(fh.read())]
print(sum(numbers))

Answer 3

h = open('file.txt')
nos = list()
for ln in h:
    fi = re.findall('[0-9]+', ln)
    for i in fi:
        nos.append(int(i))
print('Sum:', sum(nos))