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i need to convert this over to a recursive function. the program basically prints out all variations of a string.
def comb(L):
for i in range(3):
for j in range(3):
for k in range(3):
# check if the indexes are not
# same
if (i!=j and j!=k and i!=k):
print(L[i], L[j], L[k])
# Driver Code
comb([1, 2, 3])
output:
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
This might get you started:
Every permutation of a list can be made by picking a value from the list & putting it at the front of every permutation of what is left in the list after taking that value out. (Note that the "what is left" part is a smaller list than what you started with.)
When your list is small enough, it is the only permutation.
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I have a parameter that is called f.. I don't know how to code it, should I make it as a list or it needs an if loop
f = 1 if the indicator i exists in list 1
f = -1 if the indicator i exists in list 2
f = O otherwise
I will use the parameter f in a constraint with the right side equals to f
Any help?
Should any further clarification is required I am ready.
f
constraint
You can do this in one line in python:
f = 1 if i in list1 else -1 if i in list2 else 0
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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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How to get expected answer [0, 1, 2, 3] using the range function?
I tried this Python code:
print(range(4)),
result:
range(0,4)
The result is different from online playgrounds:
result 01: range(0,4)
result 02: [0, 1, 2, 3]
The general syntax of range function:
range(start, stop, step)
start : Numerical index, at which position to start(optional)
stop : Index you go upto but not include the number(position to end)(required)
step: Incremental step size(optional)
Examples
x = range(0, 6)
for n in x:
print(n) //output: 0 1 2 3 4 5
In your case you need a list containing the integers so as #Green Cloak Guy said you should use list(range(4))
do list(range(4)).
The range() function returns a generator (only produces one value at a time). You have to convert that generator to a list.
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I am looking for an efficient way to compare rows and columns of a table against each other (>= gets 1, otherwise 0) and store the result.
Example:
0.3642286 0.7945753 0.3527125
0.3642286 1 1 0
0.7945753 0 1 0
0.3527125 1 1 1
I have 21 tables with 480*480 rows and columns. What would be a proper way of generating and storing such a matrix?
All you really need is two loops.
def compare(first, second):
result = []
for x in first:
result.append([])
for y in second:
result[-1].append(1 if x >= y else 0)
result = [list(i) for i in zip(*result)]
return result
You might consider NumPy (1) if you are regularly handling large multi-dimensional arrays.
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I have 3 lists with 3 sets of integers. I want to predict the outcome numbers using a list or by whatever other means is out there.
List1 = [0 0 2 0 1 2]
List2 = [3 0 1 0 1 0]
List3 = [0 0 2 1 1 1]
Im working in the first column, I was thinking of using a double if statement something like:
if 3 is in spot 0 of any list
and 0 is in spot 0 of any list
print 2
The other thing is I want to do this for all the other columns as well.
I know this is not complete but its on my mind like this and ive been
searching around and found nothing to solve this problem.
I would greatly appeciate any response to this question.
You cant define a list delimited with space; you have to use a comma',' instead.
l1=[0,0,2,0,1,2]
l2=[3,0,1,0,1,0,]
l3=[0,0,2,1,1,1]
spot0_list = zip(l1,l2,l3)[0]
if 3 in spot0_list and 0 in spot0_list:
print 2
And, if there are too many lists, you can use set:
spot0_list = set(zip(l1,l2,l3)[0])