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write a function super_sum(A), that takes in an array of numbers A, and returns the sum of all the numbers after doubling the odd numbers and halving the even numbers.
heres my solution: how do solve this with a list comprehension
def super_sum(A):
new_sum=[]
total=0
for x in A:
if x%2==0:
x=x/2
elif x%2!=0:
x=x*2
new_sum.append(x)
return sum(new_sum)
print (super_sum([10,3,5]))
sum([x/2 if x%2==0 else x*2 for x in [10,3,5]])
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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))
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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))
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Say I have
A = [[1.0,2.3,1.1],[2.2,1.3,3.2]]
and I want to cast all of those numbers into just ints to have
A = [[1,2,1],[2,1,3]]
How do we do that in python?
Try list comprehension*2:
print([[int(x) for x in i] for i in A])
Or list comprehension + map:
print([list(map(int,i)) for i in A])
Or map+map:
print(list(map(lambda x: list(map(int,x)),A)))
Simple ways all return:
[[1,2,1],[2,1,3]]
Here's an approach using a list comprehension and map:
A = [[1.0,2.3,1.1],[2.2,1.3,3.2]]
print([list(map(int, i)) for i in A])
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I can't figure out how to do this without using complex functions, please help. this is the docstring of the code:
'''
finds all numbers in the list below a certain threshold
:param numList: a list of numbers
:threshold: the cutoff (only numbers below this will be included)
:returns: a new list of all numbers from numList below the threshold
'''
One approach
def filterList(numList, threshold):
return list(filter(lambda x: x < threshold, numList))
Another approach:
filteredList = [x for x in numList if x < threshold]
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How do I print certain elements of a list for example:
list1=["0","0","0","0","0","0","Element1","0","0","0","0"]
Is there any simple way to print only Element1 that specifies that you should not print out anything that is equal to 0.
Use a list comprehension or (as in this example) a generator expression to filter out the "0" items, and loop through the filtered list:
for item in (x for x in list1 if x != "0"):
print(item)
This prints all items that are not "0".