Currently I have a function.
def create1(n):
output = []
for i in range(n):
output.append(int(i)+1)
return output
It returns [1,2,3] whenever enter create(3). However, I want it to return [[1],[1,2],[1,2,3]]. I know there's a problem with something in my for loop but I can't figure it out.
Use range() to create lists of numbers quickly:
def create1(n):
output = []
for i in range(n):
output.append(range(1, i + 2))
return output
or, using a list comprehension:
def create1(n):
return [range(1, i + 2) for i in range(n)]
If you are using Python 3, turn the iterator returned by range() into a list first:
for i in range(n):
output.append(list(range(1, i + 2)))
Quick demo:
>>> def create1(n):
... return [range(1, i + 2) for i in range(n)]
...
>>> create1(3)
[[1], [1, 2], [1, 2, 3]]
This works in Python 2 and Python 3:
>>> def create1(n):
... return [list(range(1,i+1)) for i in range(1,n+1)]
...
>>> create1(5)
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]]
Try this:
def create(n):
output = []
for i in range(n):
output.append(range(1,i+2))
return output
print create(3)
What you really want is a list appended at every stage. So try this
def create1(n):
output = []
for i in range(n):
output.append(range(1,i+2)) # append a list, not a number.
return output
range(n) gives you a list of integers from 0 to n-1. So, at each stage (at each i), you're appending to the output a list from 0 to i+1.
