CODEWITHSUNDEEP
pythonstringfor-loopnumbers
Frixoe
Frixoe

Reputation: 85

Finding Continuing Numbers in Python

First of all, I apologize for my vague question framing, I just couldn't figure out what to call this question. I have also searched google a lot trying to find the solution but I came to no conclusion.

I currently have a whole string with numbers like these 15-5.{123434}.15-18.. I want to iterate over the string, and as soon as I encounter an opening curly brace I want to take all of numbers after that before encountering a closing brace and put them in a variable. I have thought of a way of doing this, like this:

string = '15-5.{123434}.15-18.'
result = ""

for index, letter in enumerate(string):  # For every letter with it's index,
    if letter == '{':  # If letter is '{',
        for number in string[index:]:  # Start a loop from that index,
            if number =='}':  # If number becomes '}',
                break
            result += number  #Else append the number in the result string.

However, I was wondering if there was a better way of doing this. I appreciate all of the answers. Thank you!

System: Windows 10, 32bit
Python version: 3.6.0

Upvotes: 0

Views: 253

Answers (2)

f5r5e5d
f5r5e5d

Reputation: 3741

not knocking a regex but I think I found another one-liner with .split()

[e.split('}')[0] for e in s.split('{') if '}' in e]

testing:

ss = ['','{}','{0}','1','{0}1','1{0}','{0}{2}','{0}1{2}','{0}1{2}3']                      

for s in ss:                       
    print(s,[e.split('}')[0] for e in s.split('{') if '}' in e])
 []
{} ['']
{0} ['0']
1 []
{0}1 ['0']
1{0} ['0']
{0}{2} ['0', '2']
{0}1{2} ['0', '2']
{0}1{2}3 ['0', '2']

edited to expand on list comprehensions

plenty of info available on list comprehensions, http://treyhunner.com/2015/12/python-list-comprehensions-now-in-color/ is very basic, enough to understand my answer's list comprehension

we can rewrite the comprehension with line breaks between its parts which may help reading it: 

c = [e.split('}')[0]
            for e in s.split('{')
                if '}' in e]

and we can do the whole thing in the equivalent for loop:

def in_curly(s):    
    new_list = []
    for e in s.split('{'):
        if '}' in e:
            new_list.append(e.split('}')[0])
    return new_list

c = in_curly(s)

with say:

s ='35{0}17{2}3'

either or both of the above can be stepped through in http://pythontutor.com live execution visualization mode

Upvotes: 1

inspectorG4dget
inspectorG4dget

Reputation: 114035

This is one of the few times I'll recommend a regex:

In [27]: for i in re.findall("\{\d+\}", '15-5.{123434}.15-18.'): print(i[1:-1])
123434

Of course, you could write a function to parse it yourself:

def getNums(s):
    answer = []
    curr = []
    acc = False
    for char in s:
        if char == "{":
            acc = True
            continue
        if char == "}":
            acc = False
            answer.append(''.join(curr))
            acc = False
            continue
        if acc:
            curr.append(char)

Upvotes: 4

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