The following code took me by surprise. I was hoping that I could write a function that might or might not act as a generator.
def mayGen(limit=5, asGenerator=False):
result = []
for i in range(1, limit + 1):
n = i * i
if asGenerator:
yield n
else:
result.append(n)
return result
print(mayGen(5, False))
for x in mayGen(5, True):
print(x)
But no, the output of this program is
<generator object mayGen at 0x7fa57b6ea7b0>
1
4
9
16
25
Calling mayGen with asGenerator=False is simply useless. It seems that the mere existence of a yield statement, regardless of whether it is executed, radically changes the behavior of the containing function.
So what if mayGen was actually big and complicated and I wish I could ask it to either function as a generator, or write its computed output to a file? Is there a well-traveled path here? How are people handling this sort of thing?
Just write a generator. You can always take its output and do whatever you want with it: put it in a list, write it to a file, etc. Although, if that's what you're going to be using it for a lot, you might want to write a wrapper function to do it for you, for example:
def willGen(limit=5):
for i in range(1, limit+1):
n = i * i
yield n
def willReturnList(*args, **kwargs):
return list(willGen(*args, **kwargs))
Edit: For a number of reasons mentioned in comments, this is probably not a great idea. In any event, you can write this:
def mayGen(limit=5, asGenerator=False):
def _generate(limit):
for i in range(1, limit + 1):
n = i * i
yield n
def _return(limit):
result = []
for i in range(1, limit + 1):
n = i * i
result.append(n)
return result
if asGenerator:
return _generate(limit)
return _return(limit)
Edit:
to simiplify even further, you should probably just return list(_generate(limit)) in the final line.
Related
Consider:
def itr(n):
s = 0
for i in range(0, n+1):
s = s + i * i
return s
This is a simple iterative function that I would like to change into a recursive function.
def rec(n):
import math
if n!=0:
s=n-(2*math.sqrt(n))
if s!=0:
return(s+rec(n))
else:
return(n)
else:
return n
This is my try at doing the said thing, but I cannot quite get it right.
Why does not my solution work? What is the solution?
Use:
def recursive(total, n):
if n == 0:
return total
else:
return recursive(total + n * n, n - 1)
A couple of thoughts:
This can be refactored to using only a single argument, but by supplying both the total and the current iteration count, it is easier to see how to transform the iterative approach to a recursive one.
While this function can be made recursive, it should not be, as there isn't any advantage over the iterative approach.
Let's say I have a string
S = "qwertyu"
And I want to build a list using recursion so the list looks like
L = [u, y, t, r, e, w, q]
I tried to write code like this:
def rec (S):
if len(S) > 0:
return [S[-1]].append(rec(S[0:-1]))
Ideally I want to append the last element of a shrinking string until it reaches 0
but all I got as an output is None
I know I'm not doing it right, and I have absolutely no idea what to return when the length of S reaches 0, please show me how I can make this work
(sorry the answer has to use recursion, otherwise it won't bother me)
Thank you very much!!!
There are many simpler ways than using recursion, but here's one recursive way to do it:
def rec (S):
if not S:
return []
else:
temp = list(S[-1])
temp.extend(rec(S[:-1]))
return temp
EDIT:
Notice that the base case ensures that function also works with an empty string. I had to use temp, because you cannot return list(S[-1]).extend(rec(S[:-1])) due to it being a NoneType (it's a method call rather than an object). For the same reason you cannot assign to a variable (hence the two separate lines with temp). A workaround would be to use + to concatenate the two lists, like suggested in Aryerez's answer (however, I'd suggest against his advice to try to impress people with convoluted one liners):
def rec (S):
if not S:
return []
else:
return list(S[-1]) + rec(S[:-1])
In fact using + could be more efficient (although the improvement would most likely be negligible), see answers to this SO question for more details.
This is the simplest solution:
def rec(S):
if len(S) == 1:
return S
return S[-1] + rec(S[:-1])
Or in one-line, if you really want to impress someone :)
def rec(S):
return S if len(S) == 1 else S[-1] + rec(S[:-1])
Since append mutates the list, this is a bit difficult to express recursively. One way you could do this is by using a separate inner function that passes on the current L to the next recursive call.
def rec(S):
def go(S, L):
if len(S) > 0:
L.append(S[-1])
return go(S[0:-1], L)
else:
return L
return go(S, [])
L = [i for i in S[::-1]]
It should work.
I don't know, why isn't working that code in python. The problem is, that I can get numbers, but sometimes some numbers are uniform. And I want to do 5 different numbers between 1 and 90.
class lottery:
def __init__(self):
self.price = 50
def list(self):
numbers = []
for i in range(0,5):
numbers.append(random.randint(0,90))
for i in range(1,5):
for j in range(0,i-1):
if (numbers[i]==numbers[j]):
game.list()
return numbers
game = lottery()
game.list()
Or is there any better way to solve my problem?
Thanks!
Use random.sample:
def list(self):
return random.sample(xrange(90), 5)
This is (especially for large values of 5) much more efficient than starting over every time your randomization creates a repeat, and also avoids the possibility of overflowing the stack.
First of all you should import the random module:
import random
The problem is that you are not returning the result that you get from the recursive call, therefore you are still returning the list with repeated numbers. It should be:
def list(self):
numbers = []
for i in range(0, 5):
numbers.append(random.randint(0, 90))
for i in range(1, 5):
for j in range(0, i - 1):
if (numbers[i] == numbers[j]):
return self.list() # return
return numbers
note that self is used to access to the instance of the class that calls the method. Also, don't forget to print the results:
game = lottery()
print game.list()
Note:
Don't use list as the name of a variable or method because it will hide the built-in definition of list.
I am trying to write a piece of code that will generate a permutation, or some series of characters that are all different in a recursive fashion.
def getSteps(length, res=[]):
if length == 1:
if res == []:
res.append("l")
res.append("r")
return res
else:
for i in range(0,len(res)):
res.append(res[i] + "l")
res.append(res[i] + "r")
print(res)
return res
else:
if res == []:
res.append("l")
res.append("r")
return getSteps(length-1,res)
else:
for i in range(0,len(res)):
res.append(res[i] + "l")
res.append(res[i] + "r")
print(res)
return getSteps(length-1,res)
def sanitize(length, res):
return [i for i in res if len(str(i)) == length]
print(sanitize(2,getSteps(2)))
So this would return
"LL", "LR", "RR, "RL" or some permutation of the series.
I can see right off the bat that this function probably runs quite slowly, seeing as I have to loop through an entire array. I tried to make the process as efficient as I could, but this is as far as I can get. I know that some unnecessary things happen during the run, but I don't know how to make it much better. So my question is this: what would I do to increase the efficiency and decrease the running time of this code?
edit = I want to be able to port this code to java or some other language in order to understand the concept of recursion rather than use external libraries and have my problem solved without understanding it.
Your design is broken. If you call getSteps again, res won't be an empty list, it will have garbage left over from the last call in it.
I think you want to generate permutations recursively, but I don't understand where you are going with the getSteps function
Here is a simple recursive function
def fn(x):
if x==1:
return 'LR'
return [j+i for i in fn(x-1) for j in "LR"]
Is there a way to combine the binary approach and a recursive approach?
Yes, and #gribbler came very close to that in the post to which that comment was attached. He just put the pieces together in "the other order".
How can you construct all the bitstrings of length n, in increasing order (when viewed as binary integers)? Well, if you already have all the bitstrings of length n-1, you can prefix them all with 0, and then prefix them all again with 1. It's that easy.
def f(n):
if n == 0:
return [""]
return [a + b for a in "RL" for b in f(n-1)]
print(f(3))
prints
['RRR', 'RRL', 'RLR', 'RLL', 'LRR', 'LRL', 'LLR', 'LLL']
Replace R with 0, and L with 1, and you have the 8 binary integers from 0 through 7 in increasing order.
You should look into itertools. There is a function there called permutations which does exactly what you want to achieve here.
So I'm trying to learn python on my own, and am doing coding puzzles. I came across one that pretty much ask for the best position to stand in line to win a contest. The person running the contest gets rid of people standing in odd number positions.
So for example if 1, 2, 3, 4, 5
It would get rid of the odd positions leaving 2, 4
Would get rid of the remaining odd positions leaving 4 as the winner.
When I'm debugging the code seems to be working, but it's returning [1,2,3,4,5] instead of the expected [4]
Here is my code:
def findWinner(contestants):
if (len(contestants) != 1):
remainingContestants = []
for i, contestant in enumerate(contestants, 1):
if (isEven(i)):
remainingContestants.append(contestant)
findWinner(remainingContestants)
return contestants
Am I not seeing a logic error or is there something else that I'm not seeing?
You must return the value from the recurse function to the caller function:
return findWinner(remainingContestants)
else you would return just the original value without any changes.
def findWinner(contestants):
if (len(contestants) != 1):
remainingContestants = []
for i, contestant in enumerate(contestants, 1):
if (isEven(i)):
remainingContestants.append(contestant)
return findWinner(remainingContestants) # here the value must be return
return contestants # without the return above, it will just return this value(original)
How about this:
def findWinner(contestants):
return [contestants[2**int(math.log(len(contestants),2))-1]]
I know its not what the questions really about but I had to =P. I cant just look at all that work for finding the greatest power of 2 less than contestants and not point it out.
or if you don't like the 'artificial' solution and would like to actually perform the process:
def findWinner2(c):
while len(c) > 1:
c = [obj for index, obj in enumerate(c, 1) if index % 2 == 0] #or c = c[1::2] thanks desfido
return c
you shold use
return findWinner(remaingContestants)
otherwise, of course, your list will never be updated and so your func is gonna always return containts
however, see the PEP8 for style guide on python code: http://www.python.org/dev/peps/pep-0008/
the func isEven is probably an overkill...just write
if not num % 2
finally, recursion in python isn't recommended; make something like
def find_winner(alist):
while len(alist) > 1:
to_get_rid = []
for pos, obj in enumerate(alist, 1):
if pos % 2:
to_get_rid.append(obj)
alist = [x for x in alist if not (x in to_get_rid)]
return alist
Is there a reason you're iterating over the list instead of using a slice? Doesn't seem very python-y to not use them to me.
Additionally, you might want to do something sensible in the case of an empty list. You'll currently go into an infinite loop.
I'd write your function as
def findWinner(contestants):
if not contestants:
raise Exception
if len(contestants)==1:
return contestants[0]
return findWinner(contestants[1::2])
(much as #jon_darkstar's point, this is a bit tangential to the question you are explicitly asking, but still a good practice to engage in over what you're doing)
You are missing a return at the line where you call "findWinner"