Python string from list comprehension

Question:

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?

Asked By: Reno

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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.

Answered By: underrun

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)

Answered By: bereal

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)
Answered By: hexparrot

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
Answered By: Daedalus

In Python 3.6 you could write:

markers = [(97,64),(45,84)]
result = ''.join(f'&markers={pair}' for pair in markers)
return result
Answered By: Pluto
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