Question

我可以使用它，但是在创建列表/调用列表时很糟糕。谁能将其压缩为列表形式？（不是为了做作业，已经完成了，只是为了学习机会/练习，所以不要着急。）我假设我的count变量可以放入列表中，并在for循环中和最终打印时调用。

这是一个骰子游戏，该游戏跟踪投掷2个骰子并加在一起后一个数字出现多少次。太容易了，我只是在理解列表，所以对代码的解释会很有意义，我的教科书对此主题的阐述还不够。首先发表在栈上，请原谅任何混乱的行话或规则。谢谢！

import random

count2 = 0
count3 = 0
count4 = 0
count5 = 0
count6 = 0
count7 = 0
count8 = 0
count9 = 0
count10 = 0
count11 = 0
count12 = 0

rolls = int(input("How many times would you like to roll? "))



for i in range(rolls):

    die1 = random.randint(1, 6)
    print("Roll number 1 = " + str(die1))
    die2 = random.randint(1, 6)
    print("Roll number 2 = " + str(die2))
    total = die1 + die2
    print(total)


    if total == 2:
        count2 += 1
    elif total == 3:
        count3 += 1
    elif total == 4:
        count4 += 1
    elif total == 5:
        count5 += 1
    elif total == 6:
        count6 += 1
    elif total == 7:
        count7 += 1
    elif total == 8:
        count8 += 1
    elif total == 9:
        count9 += 1
    elif total == 10:
        count10 += 1
    elif total == 11:
        count11 += 1
    elif total == 12:
        count12 += 1


print("You rolled " + str(count2) + " 2's")
print("You Rolled " + str(count3) + " 3's")
print("You Rolled " + str(count4) + " 4's")
print("You Rolled " + str(count5) + " 5's")
print("You Rolled " + str(count6) + " 6's")
print("You Rolled " + str(count7) + " 7's")
print("You Rolled " + str(count8) + " 8's")
print("You Rolled " + str(count9) + " 9's")
print("You Rolled " + str(count10) + " 10's")
print("You Rolled " + str(count11) + " 11's")
print("You Rolled " + str(count12) + " 12's")

Answer 1

您可以在此处使用列表或字典。我倾向于使用字典，我认为它最能代表您要在这里使用的稀疏数据结构（count列表的第一个元素是什么？它将始终为零，但不是吗？真的是 nothing 吗？零没有被掷过一次还是完全不能被滚出是否更有意义？）

最好将字典定义为：

counts = {}

# You can also generalize your rolling 2d6!
def roll_dice(num_dice, sides):
    total = 0
    for _ range(num_dice):
        dieroll = random.randint(1, sides)
        total += dieroll
    return total

for _ in range(rolls):
    roll = roll_dice(2, 6)
    counts.setdefault(roll, 0) += 1  # this is using dict.setdefault

for roll, count in sorted(counts.items()):
    print("You rolled {} {}s".format(count, roll))

您也可以使用collections.Counter来完成此操作。

rolls = [roll_dice(2, 6) for _ in num_rolls]
# this will generate a list like [3, 12, 6, 5, 9, 9, 7, ...],
# just a flat list of rolls.

result = collections.Counter(rolls)

Answer 2

稍微修改了您的代码，我认为如果您使用字典，那么它将易于使用和理解。

import random

# count2 = 0
# count3 = 0
# count4 = 0
# count5 = 0
# count6 = 0
# count7 = 0
# count8 = 0
# count9 = 0
# count10 = 0
# count11 = 0
# count12 = 0

count = {i: 0 for i in range(2,13)}

rolls = int(input("How many times would you like to roll? "))



for i in range(rolls):

    die1 = random.randint(1, 6)
    print("Roll number 1 = " + str(die1))
    die2 = random.randint(1, 6)
    print("Roll number 2 = " + str(die2))
    total = die1 + die2
    print(total)



    # if total == 2:
    #     count2 += 1
    # elif total == 3:
    #     count3 += 1
    # elif total == 4:
    #     count4 += 1
    # elif total == 5:
    #     count5 += 1
    # elif total == 6:
    #     count6 += 1
    # elif total == 7:
    #     count7 += 1
    # elif total == 8:
    #     count8 += 1
    # elif total == 9:
    #     count9 += 1
    # elif total == 10:
    #     count10 += 1
    # elif total == 11:
    #     count11 += 1
    # elif total == 12:
    #     count12 += 1

    count[total] += 1
# print("You rolled " + str(count2) + " 2's")
# print("You Rolled " + str(count3) + " 3's")
# print("You Rolled " + str(count4) + " 4's")
# print("You Rolled " + str(count5) + " 5's")
# print("You Rolled " + str(count6) + " 6's")
# print("You Rolled " + str(count7) + " 7's")
# print("You Rolled " + str(count8) + " 8's")
# print("You Rolled " + str(count9) + " 9's")
# print("You Rolled " + str(count10) + " 10's")
# print("You Rolled " + str(count11) + " 11's")
# print("You Rolled " + str(count12) + " 12's")

for i in range(2,13):    
    print("You rolled " + str(count[i]) + " "+i+"'s")

Answer 3

创建11个零的列表：

counts = [0] * (12 - 2 + 1)

要增加计数：

counts[total - 2] += 1

现在在一起：

import random

def roll_dice():
    die1 = random.randint(1, 6)
    die2 = random.randint(1, 6)
    total = die1 + die2

    print(f"Roll number 1 = {die1}")
    print(f"Roll number 2 = {die2}")
    print(total)

    return total

min_count = 2
counts = [0] * (12 - min_count + 1)
rolls = int(input("How many times would you like to roll? "))

for i in range(rolls):
    total = roll_dice()
    counts[total - min_count] += 1

print('\n'.join(f"You rolled {x} {i + min_count}'s"
    for i, x in enumerate(counts)))

Answer 4

我刚刚对您的代码做了一点编辑：

select ll.line_id, ll.line_value, bb.branch_unit_cd,ll.unit
  from 
(
select l.*, b.parent
  from line l
  left outer join Branch b
    on ( l.unit = b.branch_unit_cd )
) ll inner join
   Branch bb
 on ( ll.parent = bb.branch_unit_id );

LINE_ID LINE_VALUE  BRANCH_UNIT_CD  UNIT
------- ----------  --------------  ----
1       123456      FFF             EEE

Answer 5

在这种情况下，我会使用字典。或特别是defaultdict。

import random
from collections import defaultdict

roll_scores = defaultdict(int)
rolls = 10

for _ in range(rolls):
    die1 = random.randint(1, 6)
    die2 = random.randint(1, 6)
    total = die1 + die2

    print("Roll 1: ", die1)
    print("Roll 2:", die2)
    print("Total:", total)

    roll_scores[total] +=1

for k in roll_scores:
    print("You rolled {} {}'s".format(roll_scores[k], k))

但是，如果要使用列表，则概念几乎相同。将roll_scores更改为13个项目列表（0到12）：

roll_scores = [0]*13

并在最后更改打印：

for i in range(len(roll_scores)):
    print("You rolled {} {}'s".format(roll_scores[i], i))

我将如何使用列表/数组而不是多个变量来制作此Python程序？

5 个答案: