比較兩個列表並「回顧」

Question

您如何比較兩個列表並「回顧」？

我比較兩個列表的內容是這樣的：

score = 0 
for (x,y) in zip(seqA,seqB): 

    if x == y: 
     score = score +1 

    if x !=y : 
     score = score - 1 

現在我想score + 3如果以前對是比賽，所以基本上我將不得不「回頭看」一個迭代。

Answer 1

只保存最後一場比賽的結果。

score = 0 
prev = 0 

for (x,y) in zip(seqA,seqB): 

    if x == y: 
     if prev == 1: 
      score = score +3 
     else: 
      score = score +1 
     prev = 1 

    if x !=y : 
     score = score - 1 
     prev = 0 

Answer 2

可能有更直接的辦法，但是明確也不錯。
添加想法引入一個變量，告訴我們下一次匹配哪個數量。多個連續兩場比賽

score = 0 
matchPts = 1     // by default, we add 1 
for (x,y) in zip(seqA,seqB): 

    if x == y: 
     score = score + matchPts 
     matchPts = 3 

    if x !=y : 
     score = score - 1 
     matchPts = 1 

一個更復雜的獎勵規模可能有一些變化來介紹：

score = 0 
consecutiveMatches = 0 
for (x,y) in zip(seqA,seqB): 

    if x == y: 
     consecutiveMatches += 1 
     reward = 1 
     if consecutiveMatches == 2: 
      reward = 3; 
     if consecutiveMatches > 2 : 
      reward = 5; 
     if consecutiveMatches > 5 : 
      reward = 100;  // jackpot ;-) 
     // etc. 
     score += reward 
    else: 
     score -= 1 
     consecutiveMatches = 0 

Answer 3

score = 0 
previousMatch == False 
for (x,y) in zip(seqa, seqb): 
    if x==y && previousMatch: 
     score += 3 
    elif x==y: 
     score += 1 
     previousMatch = True 
    else: 
     score -= 1 
     prviousMatch = False 

Answer 4

到其他人如何做到這一點，但我寧願使用的變量名類似像「正確」，而不是看到所有的地方「x == y」...

 
# Create a list of whether an answer was "correct". 
results = [x == y for (x,y) in zip(seqA, seqB)] 

score = 0 
last_correct = False 
for current_correct in results: 
    if current_correct and last_correct: 
     score += 3 
    elif current_correct: 
     score += 1 
    else: 
     score -= 1 

    last_correct = current_correct 

print score