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I'm trying to write a program that adds only the third item in a list with dictionaries. Please help me. Here is my code so far.
# SalesManager Code
# pseudo sales list
sales = {"Meat Loaf": [50, 69, 19],
"Mineral Water": [5, 15, 10]
}
for key, value in sales.items():
total = total + value[2]
print(total)
Define total outside the loop:
total = 0
for key, value in sales.items():
total = total + value[2]
print(total)
print(total) # 29
you can use below one liner code to achieve this:
print(sum(v[2] for v in sales.values()))
You first need to declare your variable total:
sales = {"Meat Loaf": [50, 69, 19],
"Mineral Water": [5, 15, 10]
}
total = 0
for key, value in sales.items():
total = total + value[2]
print(total)
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def compute_scores(history, queries):
scores = []
for q in queries:
score = helper(history, q)
scores.append(score)
return scores
def helper(history, query, level = 1):
points = 10
for match in history:
if match[0] == query:
if len(match) == 1:
points = points
else:
for opponent in match[1:]:
opponent_score = helper(history, opponent, level)
points += int(0.2 * level * opponent_score)
level += 1
return points
history = [[1], [2, 5, 4], [3], [4], [5, 3, 1]]
queries = [2, 1, 5]
print(compute_scores(history, queries))
It passes the first test and produces the following output in a array: 17 10 16. It failed the other test cases. Not sure why. Please help!
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I am making a game where two players take turns adding a number 1 to 10 to the running total. The player who adds to get 100 wins. I stored the inputs of the players in a list, and at the end of the game, I want to display the total after every turn. The list is in the format:
allTurns = [player1move1, player2move1, player1move2, player2move2, ...]
How can I add only the first 2 elements and display them, then add only the first 4 elements and display them, and then the first 6, and so on?
for i in range(0, len(list)):
if i % 2 == 0:
print(sum(list[:i+1]))
Or
sums = []
for i in range(0, len(list)):
if i % 2 == 0:
sums.append(sums[-1] + list[i-1] + list[i])
Try this:
allTurns = [1,2,3,6,12,23]
out = []
for n in range(len(allTurns)):
if n % 2 is 1:
out.append(allTurns[n] + allTurns[n-1])
if len(allTurns) % 2 is 1:
out.append(allTurns[-1])
print(out)
I've coded it so that it could work for any situation, and not just yours.
Try this:
allTurns = [1,2,3,6,12,23,4]
out = []
for n in allTurns:
if allTurns.index(n) % 2 is 0:
i = 0
for a in range(allTurns.index(n)+2):
try:
i=i+allTurns[a]
except IndexError:
pass
out.append(i)
print(out)
This is probably as simple as it can get.
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So..I need some help for this project. I'm starting to learn a bit about python and our discussion is about File Handling. I made the correct execution but the problem here is that I need to find the sum and the average of the given loop.
def num3():
ifile = open(r'sales.txt','w')
no_days = 7
for count in range(1, no_days+1):
print('List of Sales in Php')
sales = float(input('Day' +str(count)+ ':\tPhp'))
ifile.write(str(sales)+'\n')
ifile.close()
num3()
This should work:
def num3():
with open('sales.txt','w') as fp:
no_days = 7
print('List of Sales in Php')
total = 0
for count in range(1, no_days+1):
sales = float(input('Day' +str(count)+ ':'))
total += sales
total = total
average = total/no_days
fp.write(f'Total Value: {total}\nAverage Value: {average}')
num3()
you need to accomulate the sales in a sperate variable. and then divide it by no_days:
ifile = open(r'`enter code here`sales.txt', 'w')
no_days = 7
total_sales = 0
for count in range(1, no_days + 1):
print('List of Sales in Php')
sales = float(input('Day' + str(count) + ':\tPhp'))
total_sales += sales
ifile.write(str(sales) + '\n')
ifile.close()
print(f"sum={total_sales}")
print(f"average={total_sales / no_days}")
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I don't want to use the weekday function for this as it is simplified due to the conditions.
Something like this.
weekday = {
"Mon": 0,
"Tue": 1,
"Wed": 2,
"Thu": 3,
"Fri": 4,
"Sat": 5,
"Sun": 6
}
def wd(s, k):
s = list(weekday.values())[list(weekday.keys()).index(s)]
k %= 7
result = (s + k) % 7
return result
for key, value in weekday.items():
if value == wd('Sat', 1):
print(key)
Is this what you need?
import datetime
import calendar
my_date = datetime.date(2020,5,19)
print(calendar.day_name[my_date.weekday()])
prints the name of the day..
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I just started taking classes of python and I am trying to make a code in where the value of the first month = 1 and multiply its value by 2 next month, and then by 3 next month, and 2 next month and so on. Until it reaches 6 months. I am using this code but it only gives me Month 1 = 1 which is the initial value.
P = 1
count = 12
print ("month 1: ",P)
for month in range(count-1):
if month %2 == 0:
P = P*2
else:
P = P*3
print:("month", month+2 ,":",P)
Change
print:("month", month+2 ,":",P)
to
print("month", month+2 ,":",P)
I'm not sure why python didn't complain about the colon. You can actually put anything there
weird: ("month", month+2 ,":",P)
And it won't complain. Awesome mistake, thanks!
Remember that range(count-1) will return numbers between 0 and 11 (the last number is non-inclusive)
Your logic could be like this:
(a) A counter that starts at 1 (cnt)
(b) A loop that watches for the counter to reach 6, then exits
(c) A running total (month_val or some such)
cnt = 1
month_val = 1
while cnt < 7:
month_val = month_val * cnt
print(month_val)
cnt += 1
The above assumes that you retain the new value of month_num -- but re-reading your question, you might just want to print the values 1 to 6, in which case the month_num value should just remain 1 all the time:
cnt = 1
month_val = 1
while cnt < 7:
print(month_val * cnt)
cnt += 1
You can define a dictionary for your problem like
month_values={}
for i in range(1,7):
if i == 1:
month_values['Month'+str(i)]=1
elif i%2 == 0:
month_values['Month'+str(i)]=i*2
elif i%2 == 1:
month_values['Month'+str(i)]=i*3
print(month_values)
prints
{'Month1': 1, 'Month2': 4, 'Month3': 9, 'Month4': 8, 'Month5': 15, 'Month6': 12}