如何计算字符串列表中字符的出现次数？

Question

缩进在这里是混乱的，但我想数的次数“.”出现在迷宫上，并使剩余的位数。计数似乎不起作用。当我使用bitsLeft变量时，它保持为0，而不是计算数组中出现“.”的次数

level_1 = [
"$$$$$$$$$$$$$$$$$$$$$$$$$",
"$!0......>0........#0>..$",
"$.0000.00000.0000000000.$",
"$...00.0#....0!00.0.000.$",
"$00.00.00000.0.00.0.000.$",
"$#0..........0..0.0.....$",
"$.0000.00000.00.0.0.000.$",
"$......0........0.0.0...$",
"$.000000.00000000...0.0.$",
"$.0......0...0...00.0.0.$",
"$...000000>0...0....0.0.$",
"$00.000000000000000.0.0.$",
"$...0.....0000!0........$",
"$.000.000.0>#0.0.00000.0$",
"$.....0.0.0..0.0.0..0..0$",
"$.00000.0.0..0.0.0..0.0.$",
"$.0>....0.0000.0.0..0.0.$",
"$.00000.0...P..0.0000.0.$",
"$....00.00000000.0....0.$",
"$000.00..........0.0000.$",
"$!0..00.00000000.0......$",
"$.0.000..........0.0000.$",
"$.0.0000.0000000.0.0>00.$",
"$........0#........0....$",
"$$$$$$$$$$$$$$$$$$$$$$$$$"
]

bitsLeft = level_1.count ('.')

Answer 1

您可以将sum和获取每个字符串计数的列表理解结合起来，如下所示：

sum(s.count(".") for s in level_1)

结果：

265

Answer 2

string.count()需要一个字符串。可以使用''.join(level_1)连接str.count()方法的字符串：

bitsLeft = ''.join(level_1).count ('.')