I am a python student, and I have a problem [closed] - python

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def thing():
_list = [1,1,5,6,8,3,4,6,10,23,5,12,67,3,25,2,6,5,4,3,2,1]
_list1 = [str(i) for i<=5 in lista]
return " ".join(_list1)
print(thing()))
I am new to this type of list managment, I am trying to put in _list1 only integers that are less then 5


so like the name of your function, it's created to calculate the sum of integers.
So basically, if your n=0 then your program will return 0 as result.
If not (else), it gonna give you a result of the calculation in return.
And here you want to print the interger sum of 451 which will give you the result of n % 10 + integer_sum(int(n / 10))

Related

input function : python [closed]

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counttonum = 1
countnum = input("[•] Provide A Number :")
while counttonum < countnum:
print("[", counttonum, "] : Number = ", counttonum)
counttonum +=1
I was trying to make a counting tool that counts up to the provided number from the “input()” function.
For example:
providedNumberFromInput = 5
output = 1
2
3
4
5
And it’ll stop if the provided number is reached. Please help me.

You are very close to solution. Problem is that input() returns value as string so you will need to convert it. And also if you want to include entered number use <= instead of <
while counttonum <= int(countnum):

python: list of tuples search [closed]

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Need some help here
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
if the last 6 digits are located to return the first digit.
example if finds 10,11,29,30,36,47 return 2

You can use next similair to user's approach:
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
to_find = [10,11,29,30,36,47]
print(next(n for n, *nums in num if nums == to_find))
2

You can use next with a conditional generator expression:
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
search = 11
next(first for first, *rest in num if search in rest)
# 2

I want get sum in list except null by for loop using python [closed]

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I have a list that named 'ages'. I try this.
for n in ages:
if not n:
continue
a = float(n)
sum1 += a
But that only output lastone of the list. what am I doing wrong?

A little bit change and your sum result will be well
for n in ages:
if not n:
continue
a = float(n) # move left
sum1 += a

Sum in range that meet criteria [closed]

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Trying to learn a new approach.It's the classical problem of finding the sum of all multiples of 3 and 5 below 'n'.
What I want is:
print ("The sum is: ", sum(range(1, 100)))
But with an if-statement for multiples of 3 / 5.My question:How can I include that if-statement for a one-line solution?
Thanks for your help. =)// Chris S.

This is the approach you may look into:
def thesum(the_numbers):
the_numbers = [i for i in the_numbers if i % 3 == 0 or i % 5 == 0]
return print(sum(the_numbers))
thesum(range(100))

max() function on list with an integrer [closed]

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I have a list composed of integers, and I would like to do this :
freeSushi = max(sushiPrices <= sushiPrice)
sushiPrice being an integer and sushiPrices being a list.
Any idea of how I can do this ?

You can use filter before applying max
max(filter(lambda price: price <= sushiPrice, sushiPrices)

This would work, if "something" was a list of prices less than or equal to sushiPrice:
freeSushi = max(something)
So how can we create a list of prices less than sushiPrice? Comprehend?

Use a for loop like so:
for i in sushiPrices:
if i > sushiPrice:
del sushiPrices[sushiPrices.index(i)]
freeSushi = max(sushiPrices)

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