用python计算2d列表

Question

我如何计算这个二维列表中“命中”的项目数？？

grid = [['hit','miss','miss','hit','miss'],
     ['miss','miss','hit','hit','miss'],
     ['miss','miss','miss','hit','hit'],
     ['miss','miss','miss','hit','miss'],
     ['hit','miss','miss','miss','miss']]

battleships = 0
for i in grid:
    if i == "hit":
    battleships = battleships + 1
print battleships

我知道代码是错误的，但它给出了我想做什么的想法我希望？？

谢谢

Answer 1

如果我有相当多使用2D列表的代码，我将生成一个生成器，返回2D列表中的每个元素：

def all_elements_2d(l):
    for sublist in l:
        for element in sublist:
            yield element

然后你可以用它做其他事情，比如数一数所有的“命中”字符串：

hits = sum(element == 'hit' for element in all_elements_2d(grid))

Answer 2

Transaction=[['Mango','Onion','Jar','Key-chain','Eggs','Chocolates'],
['Nuts','Onion','Jar','Key-chain','Eggs','Chocolates'],
['Mango','Apple','Key-chain','Eggs'],
['Mango','Toothbrush','corn','Key-chain','Chocolates'],
['corn','Onion','Key-chain','Knife','Chocolates']
]
count1=[['Mango',0],['Onion',0],['Jar',0],['Key-chain',0],['Eggs',0],
['Chocolates',0],['Nuts',0],['Apple',0],['Toothbrush',0],['corn',0],['Knife',0]]
for j in range(0,10):
 x=0
for i in range(0,5):
 x=x+Transaction[i].count(count1[j][0]);
count1[j][1]=x
print count1

Answer 3

使用^{}：

>>> ['hit','miss','miss','hit','miss'].count('hit')
2

>>> grid = [['hit','miss','miss','hit','miss'],
...      ['miss','miss','hit','hit','miss'],
...      ['miss','miss','miss','hit','hit'],
...      ['miss','miss','miss','hit','miss'],
...      ['hit','miss','miss','miss','miss']]
>>> [row.count('hit') for row in grid]
[2, 2, 2, 1, 1]

以及^{}：

>>> sum(row.count('hit') for row in grid)
8