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HHMathewChan
HHMathewChan

Posted on • Originally published at rebirthwithcode.tech

Python Exercise 20:Find the pattern

Question

  • Given a dictionary containing up to six phrases,
    • return a list containing the matching phrases according to the given string (p).
  • Ignore any digit that is placed after or before the given string.
  • Whether the first letter is capitalized or not,
    • if all letters of the word match the given string (p), it is valid.
  • If it does not match the given string (p) then None.

Example

  • example 1 
find_pattern({ "Phrase1": "COVID-19 is no good", "Phrase2": "COVID-18 is no good", "Phrase3": "COVID-17 is no good" }, "COVID-19") ➞ ["Phrase1 = COVID-19", "Phrase2 = None", "Phrase3 = None"]
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  • example 2 
find_pattern({ "Phrase1": "Edabit is great", "Phrase2": "Edabit is very great", "Phrase3": "Edabit is really great" }, "really") ➞ ["Phrase1 = None", "Phrase2 = None", "Phrase3 = really"]
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My solution

  • algorithm 
>>get the value of each Phrase >>separate the the value into a list by white space >>store each separated value into list the list name from phase1_list to phase6_list >>check if the given string (p) appear in the each phrase for phrase1 to phrase 6: if the given string (p) appear in the phrase value add that phrase and value into one string add that string to the final list >>return a final list
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  • code 
def find_pattern(dictionary: dict, string_p: str) -> list: prefix = "Phrase" counter = 1 final_list = [] for sentence in dictionary.values(): if string_p in sentence: final_list.append(f"{prefix}{counter} = {string_p}") else: final_list.append(f"{prefix}{counter} = None") counter += 1 return final_list
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Other solution 

def find_pattern(dict_, p): phrase = {True:p,False:"None"} phrases = [] for key,val in dict_.items(): phrases.append('{} = {}'.format(key,phrase[p in val])) return sorted(phrases)
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  • shortest version 
def find_pattern(dct, p): return sorted("{} = {}".format(d, p if p in dct[d] else None) for d in dct)
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Credit

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