print formatting in python 3

Question

I am trying to print in Python 3 and am having trouble. I have a for loop in my code that looks like this:

seq = input("enter DNA sequence to search: ")
pat = re.compile('(.{10})(ATC.{3,6}CAG)')
li = []
output_lines = [] 
for mat in pat.finditer(seq):
 x = mat.end()
 li.append(mat.groups()+(seq[x:x+10],))
for u in li:
 z = u[1] 
 A = z.count('A')
 C = z.count('C') 
 G = z.count('G') 
 T = z.count('T')
 sumbases = [A,C,G,T]
 print(sumbases)

When I print sumbases, I get this for example:

[1, 2, 3, 4]
[2, 0, 1, 4]

I am trying to format the output like this:

[1, 2, 3, 4],[2, 0, 1, 4]

Can anyone show me the problem? Thanks in advance.

• user395760: That's not the output from this code. It'll print precisely A,C,G,T and a newline per iteration. I'd prefer if you posted your real code, or at least were consistent with the examples.
• drbunsen: Thanks, sorry I didn't include the entire code. I can see how the question is unclear.
• drbunsen: @delnan i edited my post to include the entire code.
• Eric O. Lebigot: You should consider using the built-in optimized collections.Counter facility.
• smci: Seems the intent of li.append(mat.groups()+(seq[x:x+10],)) is to also capture the following ten characters, why not just add a third capture group to explicitly match them: '(.{10})(ATC.{3,6}CAG)(.{10})' . Then I think you can simply write li = re.findall(pat, seq) ...

Answer 1

You can try this:

output_lines = []
for u in li: 
 z = u[1] 
 A = z.count('A') 
 C = z.count('C') 
 G = z.count('G') 
 T = z.count('T')
 sumbases = "A,C,G,T" # I suppose you format it here differently
 y = sumbases.replace("\n"," ") # not sure why you need this
 # print(y) # don't print now, print later...
 output_lines.append(y)
print(','.join(output_lines))

EDIT for your edited question:

seq = input("enter DNA sequence to search: ")
pat = re.compile('(.{10})(ATC.{3,6}CAG)')
output_lines = [] 
for mat in pat.finditer(seq):
 x = mat.end()
 z = (mat.groups()+(seq[x:x+10],)[1]
 output_lines.append(str([z.count(a) for a in 'ACGT')]))
print(','.join(output_lines))

• user395760: Nevermind my print(y, end=',') suggestion - it adds an ugly trailing comma. This is better, although it would be even better if the whole loop could be folded into a generator expression.
• drbunsen: thanks for the help. i don't think the print statement is supported in python 3 with this syntax? when i try this i get an error?
• eumiro: @sebrowns - sorry, just put brackets around it (see my edited answer). Also, if you write 'I get an error' on stackoverflow.com, then try to include it into your comment.
• Keith: @sebrowns that's right. This example is for Python 2. But it's still not clear what your actual code is.
• drbunsen: great. i think i am starting to understand now. i get a syntax error from output_lines on the penultimate line though.