Find multiple regex in each line and skip result if one of the regex doesn't match

I have a list of variables:

variables = ['VariableA', 'VariableB','VariableC']

which I'm going to search for, line by line

ifile = open("temp.txt",'r')

d = {}

match = zeros(len(variables))
for line in ifile:
    emptyCells=0
    for i in range(len(variables)):
        regex = r'('+variables[i]+r')[:|开发者_StackOverflow中文版=|\(](-?\d+(?:\.\d+)?)(?:\))?'
        pattern_variable = re.compile(regex)
        match[i] = re.findall(pattern_variable, line)

        if match[j] == []:
            emptyCells = emptyCells+1

    if emptyCells == 0:
        for k, v in match[j]:
            d.setdefault(k, []).append(v)

The requirement is that I will only keep the lines where all the regex'es matches!

I want to collect all results for each variable in a dictionary where the variable name is the key, and the value becomes a list of all matches.

The code provided is only what I've found out so far, and is not working perfectly yet...

Can you edit your question to give an example of the source file, so we could test our solutions against it?

Anyway here's a quick hack:

from collections import defaultdict
import re

variables = ['VariableA', 'VariableB', 'VariableC']
regexes = [re.compile(r'(%s)[:|=|\(](-?\d+(?:\.\d+)?)(?:\))?' % (variable,))
           for variable in variables]
d = defaultdict(list)

with open("temp.txt") as f:
    for line in f:
        results = [regex.search(line) for regex in regexes]
        if all(results):
            for m in results:
                k, v = m.groups()
                d[k].append(v)

print d