如何將2 for循環嵌套到if語句中

Question

如何將2 for循環放入if語句中？

Ships = [ 
    ["Aircraft Carrier", 5], 
    ["Battleship", 4], 
    ["Submarine", 3],  
    ["Destroyer", 3], 
    ["Patrol Boat", 2] 
] 
ships_left = ["A","B","S","D","P"] 

if [ship for ship in ships_left name for name in Ships if name[0][0] == ship]: 
    print(name[0]) 

預期輸出：

Aircraft Carrier 

這是因爲如果他們都重複一次，然後船應等於"A"和名稱應該是["Aircraft Carrier", 5]，所以name[0][0]應該"A"。

你將如何做到這一點它的代碼獨立迭代通過兩個列表和分支給定的語句關聯兩個列表？

Answer 1

計算留第一什麼船，使用if之前測試是否有任何：

left_names = [name for name, size in Ships if name[0] in ships_left] 
if left_names: 
    print(left_names[0]) 

通過計算船舶第一剩下什麼，你可以重新使用兩種if測試結果和print()函數;否則你必須做兩次相同的計算。

你也不需要兩個循環;您只需要遍歷Ships列表並根據ships_left列表測試每個名稱。我會做ships_left一組，但是，更快的會員測試：

ships_left = {"A", "B", "S", "D", "P"} 

列表中的成員資格的測試佔據了N步（其中N是列表的長度），而在一組成員資格測試需要恆定時間（O（1））。這使得除船舶一旦它被擊沉（或放置在電路板）方便，快捷太：

ships_left.remove(name[0]) 

您可以使用next()功能與發電機的表達，如果你只需要第一匹配;這樣就避免了提取出所有的名字：

ship_left = next((name for name, size in Ships if name[0] in ships_left), None) 
if ship_left: 
    print(ship_left) 

演示：

>>> Ships = [ 
...  ["Aircraft Carrier", 5], 
...  ["Battleship", 4], 
...  ["Submarine", 3], 
...  ["Destroyer", 3], 
...  ["Patrol Boat", 2] 
... ] 
>>> ships_left = {"A", "B", "S", "D", "P"} 
>>> next((name for name, size in Ships if name[0] in ships_left), None) 
'Aircraft Carrier' 
>>> ships_left.remove('A') 
>>> next((name for name, size in Ships if name[0] in ships_left), None) 
'Battleship' 
>>> ships_left.clear() # remove all ships 
>>> next((name for name, size in Ships if name[0] in ships_left), None) is None 
True