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Keep track of items in dynamic programming(similar to Knapsack problem)

Hello I'm trying to solve this dp problem: https://vjudge.net/problem/UVA-990
I'm able to solve the initial problem result using this code below:
I used recursion and a memo table arr to optimize the code.
s=list(map(int,input().split()))
t=s[0] #seconds allowed under water
w=s[1] #w
n=int(input()) #number of treasures
depth=[-1]
gold=[-1]
time=[-1]
for i in range(3):
q=list(map(int,input().split()))
depth.append(q[0])
gold.append(q[1])
time.append(q[0]*w*3)
arr = [[-1]*(t+1) for i in range(0,(n+1))]
def maxGold(n,T):
if n==0 or T==0:
return 0
if arr[n][T]!=-1:
return arr[n][T]
if time[n]>T:
answer=maxGold(n-1,T)
else:
answer=max(maxGold(n-1,T),gold[n]+maxGold(n-1,T-time[n]))
arr[n][T]=answer
return answer
result=maxGold(n,t)
print(result)
However I have no idea how to keep track of the chosen items.
I was thinking to store all indices of chosen treasures of the maxGold() output and print them later in a loop for instance.
One approach I had was to add a paramter to the maxGold() function and append to it the indices and return two result and the indices list from the function like the following:
def maxGold(n,T,l):
if n==0 or T==0:
return 0,l
if arr[n][T]!=-1:
return arr[n][T],l
if time[n]>T:
answer=maxGold(n-1,T,l)
else:
l2=l[:]
l2.append(n)
answer=max(maxGold(n-1,T,l)[0],gold[n]+maxGold(n-1,T-time[n],l2)[0])
arr[n][T]=answer
return answer,l
result=maxGold(n,t,[])
print(result[0])
list_of_indices=result[1]
length=len(list_of_indices)
#proceed outputs
However I ran into many tuple/integer type, subscriptable,iteratable errors. from this specific line even after trying to get a round the tuple output due to several outputs :
answer=max(maxGold(n-1,T,l)[0],gold[n]+maxGold(n-1,T-time[n],l2)[0])
And honestly I'm uncertain whether this approach is the right one.
Any hints?

how we can use variables, which are described inside the loop, out side the loop in python

Below is my code. Here, I am trying to read the variable gobs(x) from an input file and then I want to use it for other calculations, eg., computing error(x). But, I found, I can read it from input file properly within the loop, but when trying to use it outside the loop, only the first data is getting transferred. For all 100 data, which I read as gobs(x) inside the loop, it is showing the value of last data only, when I am using it outside the loop.
code started below
constant = 99
x0=50
z0=5
def gsyn (x):
return (constant*z0)/(z0**2+(x-x0)**2)
with open ('Grav_H_Cyln_v3_output.txt') as finp:
lines=finp.readlines()
for line in lines:
g=float(line)
x=line
def gobs (x):
return g
print (gobs(x)) # here, gobs(x) is printing properly
def error(x):
return (gsyn(x)-gobs(x))
for i in range (1, 100, 1):
x=i
print (error(x)) # here, only the first value of gobs(x) is coming
print ('stop')

This seems like a very odd solution to what is fundamentally a very simple problem. Make gobs a dictionary so you can set or retrieve gobs[x] at will.
gobs = dict()
with open ('Grav_H_Cyln_v3_output.txt') as finp:
lines=finp.readlines()
for line in lines:
g=float(line)
gobs[line] = g
print (gobs[line])

You could try creating a vector gobs[] outside the loop, and filling it up within the loop over lines.
That should do.

Instead of reassigning the value of x on each iteration of your loop, append i to a list that is declared outside of the if-block scope.
x = []
for i in range (1, 100, 1):
x.append(i)
print(x)

Assign variable if list index out of range python error

How to pass a string to a variable if an index error is found? Consider the code:
for l1, l2 in zip(open('file1.list'), open ('file2.list')):
a=fasta1[int(l1)]
b=fasta2[int(l2)]
alignments = pairwise2.align.globalxx(a,b)
top_aln = alignments[0]
aln_a, aln_b, score, begin, end = top_aln
print aln_a+'\n'+aln_b
outfast1 = aln_a
outfast2 = aln_b
A number of these functions must be imported (pairwise2 align),
but the file.lists are single column text files with one sequence id (text and numbers) per line, that are used to extract from the fasta1 and fasta2 text files.
Basically, I want to try: each list command ( a=fasta1[int(l1)]) and if there is no error (the id is in range), do as normal (assign variables a and b for that iteration), but if NOT, assign the 'a' variable some placeholder text like 'GGG':
for l1, l2 in zip(open('file1.list'), open ('file2.list')):
try:
a=fasta1[int(l1)]
except IndexError,e:
a="GGG"
continue
try:
b=fasta2[int(l2)]
except (IndexError):
b="CCC"
continue
This code doesn't quite work (when integrated with above code), which isn't surprising given my lack of python prowess, but I don't quite know why. I actually get no text output, despite the print calls... Am I thinking about this right? If there is NO error in the index, I just want it to go on and do the pairwise alignment (with the first a and b variables) and then print some text to stdout.
Any ideas?

Python's conditional (aka ternary) expressions can one-line this for you. They're often criticized for lack of readability, but I think this example reads well enough.
a = fasta1[int(l1)] if int(l1) < len(fasta1) else "GGG"

You don't need continue, because it will skip that iteration of the loop. Consider the following:
for l1, l2 in zip(open('file1.list'), open ('file2.list')):
a = 'GGG'
b = 'CCC'
try:
a = fasta1[int(l1)]
b = fasta2[int(l2)]
except IndexError:
pass

Python. Project Euler Q35. Got a solution but I don't understand why other method doesn't work.

def rot_dig(x):
y=''
output=[x]
listing=list(x)
for i in range(1,len(x)):
listing.append(listing[0])
del(listing[0])
for i in listing:
y=y+i
output.append(y)
y=''
return output
import math
def prime_is(x,prime):
for m in prime:
if m&lt=math.sqrt(x):
if x%m==0:
return False
else:
return True
prime=[2]
for x in range(3,1000000):
if prime_is(x,prime):
prime.append(x)
primestr=[]
for x in prime:
primestr.append(str(x))
sums=0
for x in primestr:
count=0
for y in rot_dig(x):
if y in primestr:
count+=1
if count==len(x):
sums+=1
else:
for y in rot_dig(x):
if y in primestr:
primestr.remove(y)
print sums
When run with the bold code the solutions miss the final rotation. So if it looks at say 1193, it includes 1193, 3119, 9311 but not 1931. I have spent a while trying to work out why but I don't get it.
I have since edited the code to make it much quicker and solved the problem I had by simply removing the block of code, but I can't understand why it happens since surely that block of code will only be executed on non circular primes.

It's probably because your outer loop is for x in primestr: and the marked code removes items from primestr. You don't want to change primestr while looping over it that way. You could use a loop like while i < len(primestr) instead.
Some other improvements would be to compute sqrt outside the loop; to use a list comprehension instead of a loop to create primestr; and especially to use string slicing in rot_dig, it's way more complicated than it needs to be.

python, how to write an iterative function

I am quering a database for some paramaters which depend on a attribute called count! count can be incremented incase the 1st query does not return anything. Here is a sample code
sls = {(213.243, 55.556): {}, (217.193, 55.793): {}, (213.403, 55.369): {}}
for key in sls.keys:
if not sls[key]:
ra, dec = key[0], key[1]
search_from_sourcelist(sl, ra,dec)
count = 1
def search_from_sourcelist(sl, ra,dec):
dist = count/3600.0
sls[(ra,dec)] = sl.sources.area_search(Area=(ra,dec,dist))
return
Incase i run the method search_from_sourcelist, and it doesnt return anything, i would like to increment count, and do the query again. This is to be done for all keys in sls dictionary, untill all the keys have a value!!

Here is the most fundamental recursive function
def countdown(n):
if n == 0:
return "Blastoff"
else:
print "T minus %s" % n
return countdown(n-1)
You will notice that countdown returns itself with a modified argument, in this case n but -1, so if you actually followed this all the way through you would get
(-> indicates a call)
countdown(5) -> countdown(4) -> countdown(3) -> countdown(2) -> countdown(1) -> countdown(0) #stop
So now you understand what a recursive function looks like you realize you never actually return a call of your own function, thus your code is not recursive
We use recursion because we want to boil a task down to its simplest form then work from there, so a good example of this would be the mcnuggets problem. So you need to tell us what you are trying to achieve and how it can be made a smaller problem (or more importantly why.) Are you sure you cannot do this iteratively? remember you don't want to blow your stack depth because python is NOT tail recursive by standard

Recursion is useful when you find a way to reduce the initial problem to a "smaller version of itself".
The standard example is the factorial function
def fac(n):
return n * fac(n-1) if n > 1 else 1
Here you reduce the problem of calculating the factorial of n to calculating the factorial of n-1.
In your code there is no such "reduction". You just increment a value and start the same problem over again. Thus, I recommend you solve it iteratively.

I'm not sure that you need a recursive algorithm for this.
Incase i run the method search_from_sourcelist, and it doesnt return anything, i would like to increment count, and do the query again. This can be done with a while loop as follows:
for key, value in sls.iteritems():
if not value:
ra, dec = key[0], key[1]
count = 1
while not search_from_sourcelist(sls, ra, dec):
count += 1
But if you really do want to do this recursively, you can do it as follows, leave a comment and I'll write it up.
Further, you should look into your search_from_sourcelist function, as it always returns None