我能让这个算法更有效率吗?

Question

这是大学作业。我必须模拟10次随机骰子滚动(1 - 6)，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滚动的次数。我的输出必须像这样:

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

我的代码是这样的:

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有效，甚至简化它。

Answer 1

这类任务的标准工具是来自标准库的collections.Counter。它是专门为这项任务而建造的。

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

---

如果由于某种原因您不允许使用collections.Counter()，那么您的实现就像它可以一样好，只有一个例外-dict是比list更好的数据结构用于滚动计数(实际上，collections.Counter是dict的子类)。这使您能够使用setdefault()和get(key, default)，从而减少了几行代码:

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]}")

不是循环一次形成一个滚动列表，然后再循环一次计数每一个，你可以这样做，循环一次，滚动和计数在同一个循环。你也不需要排序或检查计数是否存在。

Answer 3

由于值的范围如此有限，因此不需要使用任何花哨的集合来跟踪它们，当然也不需要对输入进行排序。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