我可以让这个算法更有效吗？

Question

这是大学的作业。 我必须模拟 10 个随机掷骰子（在 1 - 6 之间）并将它们放在一个列表中，然后创建另一个列表，该列表将保留每个值的滚动次数。 我的输出必须是这样的：

   Your rolls: [6, 2, 3, 4, 5, 6, 3, 1, 5, 2]
   Number of rolls:
   1 -> 1
   2 -> 2
   3 -> 2
   4 -> 1
   5 -> 2
   6 -> 2

这是我的代码结果：

from random import randint
rolls = [randint(1, 6) for _ in range(10)]
roll_count = []

print(f"Your rolls: {rolls}")
print("Number of rolls:")

rolls.sort()

for i in rolls:
    # Puts the amount of times each number (1-6) rolled inside roll_count
    roll_count.append(rolls.count(i))

    # Removes the duplicates so that the count doesn't register again
    while rolls.count(i) > 1:
        rolls.remove(i)

for n in range(1, 7):
    # Checks if the possible rolls (1-6) have been rolled, if not place the count as 0
    # in the sorted index inside roll_count
    if n not in rolls:
        roll_count.insert(n - 1, 0)

    print(f"{n} -> {roll_count[n - 1]}")

它工作正常，但我想知道我是否可以使它更高效甚至简化它。

Answer 1

这种任务的标准工具是来自标准库的collections.Counter 。 它专为这项任务而设计。

from collections import Counter
from random import randint

rolls = [randint(1, 6) for _ in range(10)]
roll_count = Counter(rolls)

print(f"Your rolls: {rolls}")
print("Number of rolls:")

for n in range(1, 7):
    print(f"{n} -> {roll_count[n]}")

Your rolls: [2, 5, 3, 3, 3, 2, 5, 5, 2, 6]
Number of rolls:
1 -> 0
2 -> 3
3 -> 3
4 -> 0
5 -> 3
6 -> 1

---

如果由于某种原因不允许您使用collections.Counter() ，那么您的实现就尽可能好，只有一个例外 - dict是比滚动计数list更好的数据结构（确实, collections.Counter是dict的子类）。 这使您可以使用setdefault()和get(key, default) ，这减少了几行代码：

from random import randint
rolls = [randint(1, 6) for _ in range(10)]
roll_count = {}

print(f"Your rolls: {rolls}")
print("Number of rolls:")

for i in rolls:
    roll_count.setdefault(i, 0)  # does nothing if element is already present
    roll_count[i] += 1

for n in range(1, 7):
    print(f"{n} -> {roll_count.get(n, 0)}")

Answer 2

from random import randint
rolls = []
roll_count = [0] * 6
  
for i in range(10):
    r = randint(1,6)
    rolls.append(r)
    roll_count[r-1] = roll_count[r-1] + 1

print(f"Your rolls: {rolls}")
print("Number of rolls:")

for n in range(1, 7):
    print(f"{n} -> {roll_count[n - 1]}")

不是循环一次以形成滚动列表然后再次循环以计算每个列表，您可以通过这种方式循环一次并滚动以及在同一循环中计数。 而且您也不需要排序或检查计数是否存在。

Answer 3

from random import randint
rolls = [randint(1, 6) for _ in range(10)]

print(f"Your rolls: {rolls}")
print("Number of rolls:")

for i in range(1,7):
    print(f"{i} -> {rolls.count(i)}")

Answer 4

由于您的值范围如此有限，因此无需使用任何花哨的集合来跟踪它们，当然也无需对输入进行排序。 一个简单的list就可以了，初始化为全零。

from random import randint
rolls = [randint(1, 6) for _ in range(10)]
roll_count = [0] * 6

print(f"Your rolls: {rolls}")
print("Number of rolls:")

for i in rolls:
    # Puts the amount of times each number (1-6) rolled inside roll_count
    roll_count[i - 1] += 1

for n in range(1, 7):
    print(f"{n} -> {roll_count[n - 1]}")


Your rolls: [4, 1, 3, 3, 4, 3, 4, 5, 2, 2]
Number of rolls:
1 -> 1
2 -> 2
3 -> 3
4 -> 3
5 -> 1
6 -> 0