比較兩個字符串的相似性在Python

Question

我想有以下幾點：

Input: ("a#c","abc") 
Output: True 

Input:("a#c","abd") 
Desired Output: False 
Real Output: True 

所以該函數返回true，如果兩個字符串的長度相同，如果他們通過＃字符，它表示只有不同爲隨機字符。 如果不是，我希望它返回False。

該功能應該改變什麼？

def checkSolution(problem, solution): 

    if len(problem) != len(solution): 
     return False 

    if len(problem) == len(solution): 
     for i, c in zip(problem, solution): 
      if i != c: 
       return False 

      if i == c or "#" == c: 
       return True 

print (checkSolution("a#c","abc")) 

print (checkSolution("a#c","abd")) 

Answer 1

現在你只是測試長度和首字符匹配。

for i, c in zip(problem, solution): 
    if i != c: 
     # that's the first set of chars, but we're already returning?? 
     return False 

    if i == c or "#" == c: 
     # wildcard works here, but already would have failed earlier, 
     # and still an early return if it IS true! 
     return True 

相反，你需要經過整個字符串，返回的結果，或使用all來爲你做它。

if len(problem) == len(solution): 
    for p_ch, s_ch in zip(problem, solution): 
     if p_ch == "#": # wildcard 
      continue # so we skip testing this character 
     if p_ch != s_ch: 
      return False # This logic works now that we're skipping testing the "#" 
    else: # if we fall off the bottom here 
     return True # then it must be equal 
else: 
    return False 

或一條線：

return len(problem) == len(solution) and \ 
     all(p_ch==s_ch or p_ch=="#" for p_ch, s_ch in zip(problem, solution) 

，或者如果你真的瘋了（讀：你喜歡的正則表達式的方法太多了），你可以這樣做：

def checkSolution(problem, solution): 
    return re.match("^" + ".".join(map(re.escape, problem.split("#"))) + "$", 
        solution) 

Answer 2

您只檢查第一個字符。如果第一個字符相同或者它是#，則不應返回True，但應繼續查找第一個不匹配項，並僅在for循環外返回True。

的第二個問題是，在你的測試用例變量c是永遠'#'，因爲i是problem一個字符，而c是solution一個字符。

def checkSolution(problem, solution): 
    if len(problem) != len(solution): 
     return False 
    for i, c in zip(problem, solution): 
     if i != '#' and c != '#' and i != c : 
      return False 
    return True 

Answer 3

正如在評論中指出的那樣，您的縮進會被忽略，應該修正。

if len(problem) == len(solution): 
    # in the line below, 
    # 'i' will contain the next char from problem 
    # 'c' will contain the next char from solution 
    for i, c in zip(problem, solution): 
     # in this line, if they're not equal, you return False 
     # before you have a chance to look for the wildcard character 
     if i != c: 
      return False 
     # ...and here you'd fail anyway, because you're testing 
     # the character from the solution string against the wildcard... 
     if i == c or "#" == c: 
      return True 
# ...while in your test, you pass the wildcard in as part of the problem string. 
print (checkSolution("a#c","abc")) 

Answer 4

您的功能的一行版本：

def check_solution(problem, solution): 
    return (len(problem) == len(solution) and 
      all(ch==solution[i] for i, ch in enumerate(problem) if ch != '#')) 

測試：

>>> check_solution("a#c", "abc") 
True 
>>> check_solution("a#c", "abd") 
False