how to compare a string with items in a list , character by character in python

Question

I have a list and I want to compare user input with that list , but character by character. for example user can only input some character and leave the rest by dots. (for example : V...r..n ) How do I can compare strings character by character, and if it only includes all characters by user input (skip the dots)

list1 = ["Vaporeon", "Jolteon", "Flareon", "Espeon", "Umbreon", "Leafeon", "Glaceon", "Sylveon"]
s = input()  # for example "V...r..n"
for i in list1:
    # if s include exactly characters in i (skip the dots)
        print(i)

nosklo · Accepted Answer · 2018-07-14 12:29:15Z

2

You can use regular expressions:

import re

list1 = ["Vaporeon", "Jolteon", "Flareon", "Espeon", "Umbreon", "Leafeon", "Glaceon", "Sylveon"]
s = input()  # for example "V...r..n"

re_s = re.compile(''.join('.' if ch == '.' else re.escape(ch) for ch in s) + '$')
for i in list1:
    if re_s.match(i):
        print(i)

EDIT: Another option that seems missing from the other answers is to use itertools.zip_longest:

from itertools import zip_longest

list1 = ["Vaporeon", "Jolteon", "Flareon", "Espeon", "Umbreon", "Leafeon", "Glaceon", "Sylveon"]
s = input()  # for example "V...r..n"
for i in list1:
    if all(c1 == c2 for c1, c2 in zip_longest(i, s) if c2 != '.'):
        print(i)

edited Jul 14, 2018 at 12:29

answered Jul 14, 2018 at 10:56

nosklo

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Comments

jpp · Accepted Answer · 2018-07-14 15:02:53Z

1

Here's one way. There are two steps. First make a dictionary mapping indices to relevant characters. Then check for equality for those indices.

L = ["Vaporeon", "Jolteon", "Flareon", "Espeon", "Umbreon", "Leafeon", "Glaceon", "Sylveon"]

s = input()
d = {k: v for k, v in enumerate(s) if v != '.'}

for item in L:
    if (len(s) == len(item)) and all(item[k] == v for k, v in d.items()):
        print(item)

# V...r..n
# Vaporeon

edited Jul 14, 2018 at 15:02

answered Jul 14, 2018 at 10:56

jpp

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1 Comment

RoadRunner Over a year ago

I guess you could have d outside the loop here, so it doesnt get created everytime on each iteration.

politinsa · Accepted Answer · 2018-07-14 10:57:18Z

0

There are several ways to solve it, with regex, with the solution of jpp, or with this one:

list1 = ["Vaporeon", "Jolteon", "Flareon", "Espeon", "Umbreon", "Leafeon", "Glaceon", "Sylveon"]
s = input()  # for example "V...r..n"
for word in list1:
    l = len(word)
    c = 0
    flag = True
    while c < l and flag:
        if s[c] != '.':
            if s[c] != word[c]:
                flag = False
        c += 1
    if flag:
        print(word)

answered Jul 14, 2018 at 10:57

politinsa

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Comments

Koyaanis · Accepted Answer · 2018-07-14 10:59:10Z

-1

There are some hard looking solutions here. I found this to be the easiest and working solution.

list1 = ["Vaporeon", "Jolteon", "Flareon", "Espeon", "Umbreon", "Leafeon", "Glaceon", "Sylveon"]
s = input()  # for example "V...r..n"
matchedList = []
for i in list1:
    inputInList = True
    for letter in s:
        if not letter in i:
            inputInList = False
            break
    if inputInList:
        matchedList.append(i)

print(matchedList)

answered Jul 14, 2018 at 10:59

Koyaanis

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2 Comments

politinsa Over a year ago

Did you say "working" solution ? Because it doesn't with V...r..n

Koyaanis Over a year ago

Weird, it works well here. Do you literally type the dots?

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