#TASK ONE
# LISTS #
import sys
import time
cows = []
val = []
num4 = 0
literd = []
test1 = []
val1 = []
day = []
end = []
num2 = []
days = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"]
daya = ["1234567"]
# COW ID VALIDATION #
h = int(input("How many cows do you have? "))
for x in range(h):
Id = int(input("Enter the 3 digit code of your cows: "))
if len(str(Id)) != 3:
print("Sorry, the code ID can only be 3 digit numbers!")
sys.exit()
elif Id in cows:
print("The cow ID is already in!")
sys.exit()
else:
cows.append(Id)
###################
# NUMBER OF COW LITERS #
print(" ")
dayz = list(days)
cowz = list(cows)
print(" ")
print("In reality a cow can get milked 25 liters a day, beware!")
for day in days:
print("Day: ", dayz[0])
cowz = list(cows)
dayz.pop(0)
for cow in cows:
print("Cow: ", cow)
test1.append(float(input("How many liters did you milk the cow at AM? ")))
test1.append(float(input("How many liters did you milk the cow at PM? ")))
val = list(test1)
########################
#ROUNDING NUMBERS AND ADDING THE LITTERS OF COW MILKED#
val = [ round(elem, 1) for elem in test1 ]
kek = list()
total = sum(map(len, daya)) * h
for x in range(total):
num1 = val[0] + val[1]
num2.append(num1)
kek.append(num1)
for x in range(2):
val.pop(0)
num3 = list(num2)
########################################################
# PRINTS OUT THE RESULTS #
print(" ")
print("A table will print out in just a second showing the results.")
cowz = list(cows)
print(" ")
for day in days:
print("Day: ", day)
for cow in cows:
print("Cow: ", cow, end = " ")
print("liters: ", num2[0])
num2.pop(0)
########################################################
cow1 = list()
#TASK TWO
# PRINTING OUT THE TOTAL LITERS OF MILK THIS WEEK#
weekly = list()
num5 = len(num3)
for x in range(num5):
num4 += num3[0]
num3.pop(0)
print("Total liters of milk this week", num4)
#################################################
# ADDING ALL THE LITERS FOR EACH COW #
for day in days:
cowz = list(cows)
for cow in cows:
cow1.append(cowz[0])
cow1.append(kek[0])
cowz.pop(0)
kek.pop(0)
##################################################
cow2 = list(cow1)
cow3 = list(cows)
cow_val = []
cow_val1 = []
print(cows, cow3)
for x in range(h):
current = cow3[0]
while current in cow3:
cow_val.append(cow1[cow1.index(current)])
cow1.pop(cow1.index(current))
cow1.pop(cow1.index(current)+1)
cow_val1.append(sum(cow_val))
cow3.pop(0)
我和我的朋友试图找出问题所在。我们找不到任何其他可能的解决方案，我真的需要这个帮助。这是打印出来的错误
Traceback (most recent call last):
File "/Users/Datboi/Documents/test.py", line 98, in <module>
cow1.pop(cow1.index(current)+1)
ValueError: 123 is not in list
我的问题是最后几行。我不知道如何解决这个问题。任何帮助将不胜感激：D。
如果您尝试运行代码并碰巧发现一些错误或任何其他改进，请提出建议。我想尝试让我的代码更短一些，所以是的。
＃更新编辑。
for x in range(h):
xo = cow3[0]
while xo in cow3:
i += 1
if x == xo:
sum1 = sum1 + cow1[i]
cow_val1.append(sum1)
cow3.pop(0)
由于某些原因，它不会发生任何事情，因此无效。就像python被打破一样。
答案 0 :(得分：0)
所以请尝试改变这样：
cow1.pop(cow1.index(current)+1)
cow1.pop(cow1.index(current))
Expl：
在删除第一个元素的代码中，数组的索引会发生变化。然后，当您尝试删除下一个元素时，索引是不同的。所以首先删除下一个元素并尝试删除当前的elemnet
Ed：
for x in range(h):
current = cow3[0]
while current in cow3:
cow_val.append(cow1[cow1.index(current)])
# Changes
cow1.pop(cow1.index(current)+1)
cow1.pop(cow1.index(current))
# Change here also
cow3.pop(0)
cow_val1.append(sum(cow_val))
UP：
c = [333, 3.0, 334, 7.0, 333, 11.0, 334, 15.0, 333, 10.0, 334, 5.0, 333, 9.0,
334, 13.0, 333, 17.0, 334, 3.0, 333, 7.0, 334, 11.0, 333, 15.0, 334, 10.0] # I think this one you are finally getting
cows = [333, 334] # your list of cows
d = {}
for i in cows:
d[i] = []
for i in range(len(c)):
if i % 2 != 0:
d[c[i - 1]].append(c[i])
print(d)
o / p：{333：[3.0,11.0,10.0,9.0,17.0,7.0,15.0]，334：[7.0,15.0,5.0,13.0,3.0,11.0,10.0]}
print(sum(d[333])) # total milk of cow 333
o / p：72.0
答案 1 :(得分：0)
如果我没弄错的话：
cow2 = list(cow1)
cow3 = list(cows)
cow_val = []
cow_val1 = []
print(cows, cow3)
for x in range(h):
current = cow3[0]
while current in cow3:
cow_val.append(cow1[cow1.index(current)])
cow1.pop(cow1.index(current))
cow1.pop(cow1.index(current)+1)
cow_val1.append(sum(cow_val))
cow3.pop(0)
为什么你在这里弹出两次？如果这是无意的，您可能需要更仔细地查看您的代码。
您可能需要考虑重新组织信息：
{"cow_idhere":{"sunday":{"AM":0, "PM":0}, \
"monday":{"AM":0, "PM":0}, \
"tuesday":{"AM":0, "PM":0}, \
"wednesday":{"AM":0, "PM":0}, \
"thursday":{"AM":0, "PM":0}, \
"friday":{"AM":0, "PM":0}, \
"saturday":{"AM":0, "PM":0}},
"nextcow_id": ... }
具有访问权限以获取信息。 （你的部分） 我不认为需要经常弹出并附加如此多的奶牛。我的意思是变量。 示例
def get_MilkForDay(cow_data = None, day=""):
def get_MilkForWeek(cow_data = None):
最后一节（上文）也没有描述它的含义。究竟该怎么做？
答案 2 :(得分：0)
cow1.pop(cow1.index(current))
cow1.pop(cow1.index(current)+1)
这是你的错误。你正在为你正在做的事情做太多的流行音乐。停止弹出奶牛！给他们小费！
第一行：你正在做的是寻找值＆＃34; 333＆＃34;的索引。让我们说这个列表有[＆＃34; 333＆＃34;，＆＃34; 1＆＃34;]。弹出第一个元素，现在列表是[＆＃34; 1＆＃34;]。
第二行：寻找＆＃34; 333＆＃34; - 错误，列表是[＆＃34; 1＆＃34;]
我强烈建议您访问信息并计算，而不是不断更改您的信息。使用一些功能，以便您的基本信息保持不变（在某些情况下，它赢了。） 将信息链接在一起，而不是将它们分开并始终重新组合信息。 EG天和大日
days = ["monday", ... "sunday"]
#access days by days[<number>]
#days[0] = "monday"
#days[1] = "tuesday" ... #days[6]="sunday"
你也可以.append（）！
milk_Evening = input("milk evening")
milk_Day = input("milk day")
days[0].append(milk_Evening) # day[0] [ 0 , 1, 2]
days[0].append(milk_Day) #now it looks like [monday, 1, 3]
#or
#days[0].append([milk_Evening, milk_Day])
星期一挤奶多少钱？ for day_data in days： 如果day_data [0] ==＆＃34;星期一＆＃34;： total_milk = day_data [1] + day_data [2]
注意没有POPPING。收集信息。然后访问信息。没有数据操作。信息从输入时保持不变。
答案 3 :(得分：0)
# ADDING TOTAL LITERS FOR EACH COW #
cow2 = list(cow1)
cow3 = list(cows)
cowval = []
cow_val1 = []
for cow in cows:
cow2 = list(cow1)
i = 0
total1 = 0
while cow3[0] in cow2:
if cow3[0] == cow2[0]:
i += 1
cowval.append(cow2[i])
cow2.pop(0)
cow2.pop(0)
i = 0
elif cow3[0] != cow2[0]:
cow2.pop(0)
cow2.pop(0)
i = 0
cow3.pop(0)
cow_val1.append(sum(cowval))
cowval.clear()