find numbers starting certain number with python

Question

i need to find years of dates of births only starting with 18xx and 19xxfrom string

i'm using regex to solve task

i have testing testbirtdays = 'ABCDEFG 01.19.1701 1801 02.18.1901 2001'

def getNumbers(str): 
    array = re.findall(r'[0-9]+', str) 
    return array

i can use this function but output will be:

getNumbers(testbirtdays)

#['01', '19', '1701', '1801', '02', '18', '1901', '2001']

my function can't do 2 things:

i need numbers only starting wtih 18 and 19
i need only 4x numbers to get only years and ignore months/days

so i need output like:

#['1801','1901']

Wiktor Stribiżew · Accepted Answer · 2020-04-02 08:35:02Z

3

You may use

r'(?<![0-9])1[89][0-9]{2}(?![0-9])'

Or, with word boundaries:

r'\b1[89][0-9]{2}\b'

See the regex demo #1 and regex demo #2.

Regex details:

(?<![0-9]) - no ASCII digit allowed immediately on the left
\b - a word boundary
1 - a 1 digit
[89] - 8 or 9
[0-9]{2} - two ASCII digit
(?![0-9]) - no ASCII digit allowed immediately on the right or
\b - a word boundary

See the Python demo:

import re

def getNumbers(s): 
    return re.findall(r'(?<![0-9])1[89][0-9]{2}(?![0-9])', s) 

testbirtdays = 'ABCDEFG 01.19.1701 1801 02.18.1901 2001'
print(getNumbers(testbirtdays)) # => ['1801', '1901']

answered Apr 2, 2020 at 8:35

Wiktor Stribiżew

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Comments

kederrac · Accepted Answer · 2020-04-02 08:44:05Z

2

here is one way :

import re

re.findall(r'\b18\d{2}\b|\b19\d{2}\b', testbirtdays)

output:

['1801', '1901']

edited Apr 2, 2020 at 8:44

answered Apr 2, 2020 at 8:35

kederrac

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Comments

azro · Accepted Answer · 2020-04-02 08:35:04Z

1

You need a more specific regex like 1[8-9][0-9]{2} : a 1, then one of 89 then 2 digits

You can also do (?:18|19)[0-9]{2} start with 18or 19 then 2 other digits

def getNumbers(value):
    return re.findall(r'1[8-9][0-9]{2}', value)

r = getNumbers('ABCDEFG 01.19.1701 1801 02.18.1901 2001')
print(r)  # ['1801', '1901']

answered Apr 2, 2020 at 8:35

azro

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Comments

Shubham Sharma · Accepted Answer · 2020-04-02 08:35:51Z

1

Try this:

def get_years(str):
    return re.findall(r"((?:18|19)\d{2})\b", str)

print(get_years(testbirtdays))

Output:

['1801', '1901']

answered Apr 2, 2020 at 8:35

Shubham Sharma

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Comments

purpin · Accepted Answer · 2020-04-02 09:37:57Z

1

test = 'ABCDEFG 01.19.1701 1801 02.18.1901 2001'
pattern = r'1[89]\d{2}'    
re.findall(pattern, test)

The pattern looks for 1 followed by 8 or 9, and 2 more digits.

Output:

['1801', '1901']

edited Apr 2, 2020 at 9:37

answered Apr 2, 2020 at 8:37

purpin

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6 Comments

The fourth bird Over a year ago

The | in the character class [] does not mean OR but a pipe char

purpin Over a year ago

But this is the output I got : ['1801', '1901']

The fourth bird Over a year ago

You get that output, but it might also match 1|01

purpin Over a year ago

Ah, right. Now I get it - [] by itself does the OR part, so it should just be [89].

The fourth bird Over a year ago

Exactly, but now you will still get partial matches in for example 55180155 which I don't think is the desired result. The answer provided by Wiktor Stribiżew shows how that can be prevented.

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