Python Function that Match a Given Template

Question

Say I have a template and I have a dictionary containing words. I want to create a function that returns a list of words that match the given template. To illustrate this:

template = 'W**ER'
dictionary = {'apple': '', 'water': '', 'weber': '', 'tiger': '', 'elder': '', 'rover': '', 'waver': ''}


Desired output:
output_list = ['water', 'weber', 'waver']

I'm not entirely sure how to create the function and I would totally appreciate any help. Thanks!

Answer 1

You can use regular expression to find a matched keys with the templete.

My code find all the matched keys, and it is case-insensitive code, as follows:

import re

dictionary = {'apple': '', 'water': '', 'weber': '', 'tiger': '', 'elder': '', 'rover': '', 'waver': ''}

output = []
for key in dictionary.keys():
    matched = re.match("[Ww]+..[Ee]+[Rr]+$", key) # case-insensitive, 5 letters
    if matched:
        output.append(key)

print(output)
#['water', 'weber', 'waver']

For more information about regular expressions with examples, you can see: https://docs.python.org/3/library/re.html

Answer 2

As all the answers contain '.' as any symbol, although it may be better to use any letter instead. Also note having start of string and end of string. That means that waterdasdadad won't proceed

Here is my code

import re

template = r'^w\w\wer$'  # means start of string, w, any alphabet char twice, e, r, end of string
regex = re.compile(template)
dictionary = {'apple': '', 'water': '', 'weber': '', 'tiger': '', 'elder': '', 'rover': '', 'waver': ''}

output_list = list()

for key in dictionary:
    if regex.fullmatch(key):  # use key.lower() for case insensitivity
        output_list.append(key)

print(output_list)

Answer 3

Try to use regex:

import re
ptrn = re.compile("w..er")
list(filter(lambda key: re.match(ptrn, key), dictionary))