在python中计算2d列表

Question

如何计算此 2d 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

使用 list.count ：

>>> ['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]

和sum ：

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

Answer 2

如果我的代码经常使用 2D 列表，我会创建一个生成器来返回 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，比如计算所有的"命中"字符串：

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

Answer 3

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