I am trying to get this string as a result:
"&markers=97,64&markers=45,84" From the Python code below:
markers = [(97,64),(45,84)] result = ("&markers=%s" %x for x in markers) return result How do I do this as the below does not give me the actual string?
5 Answers
You need to join your string like this:
markers = [(97,64),(45,84)] result = ''.join("&markers=%s" % ','.join(map(str, x)) for x in markers) return result UPDATE
I didn't initially have the ','.join(map(str, x)) section in there to turn each tuple into strings. This handles varying length tuples, but if you will always have exactly 2 numbers, you might see gatto's comment below.
The explanation of what's going on is that we make a list with one item for each tuple from markers, turning the tuples into comma separated strings which we format into the &markers= string. This list of strings is then joined together separated by an empty string.
In Python 3.6 you could write:
markers = [(97,64),(45,84)] result = ''.join(f'&markers={pair}' for pair in markers) return result While the first answer is doing what's expected, I'd make it a bit more "pythonic" by getting rid of map and nested expressions:
def join(seq, sep=','): return sep.join(str(i) for i in seq) result = ''.join('&markers=%s' % join(m) for m in markers) (if that's for urls like it seems, you can also take a look at urllib.urlencode)
Here's another approach that hopefully makes the intent the most clear by specifying the location of each of your values explicitly:
markers = [(97,64),(45,84)] print ''.join('&markers=%s,%s' % pair for pair in markers) Try creating an empty string adding to it then removing the last comma
result = ''
for i in a: result+='&markers' for j in i: result += str(j) + ',' result = result[:len(result)-1] return result 
