Concatenating strings more efficiently in Python

Question

I've been learning Python for a couple of months, and wanted to understand a cleaner and more efficient way of writing this function. It's just a basic thing I use to look up bus times near me, then display the contents of mtodisplay on an LCD, but I'm not sure about the mtodisplay=mtodisplay+... line. There must be a better, smarter, more Pythonic way of concatenating a string, without resorting to lists (I want to output this string direct to LCD. Saves me time. Maybe that's my problem ... I'm taking shortcuts).

Similarly, my method of using countit and thebuslen seems a bit ridiculous! I'd really welcome some advice or pointers in making this better. Just wanna learn!

Thanks

json_string = requests.get(busurl)
the_data = json_string.json()
mtodisplay='220 buses:\n'
countit=0
for entry in the_data['departures']:
    for thebuses in the_data['departures'][entry]:
        if thebuses['line'] == '220':
            thebuslen=len(the_data['departures'][entry])
            print 'buslen',thebuslen
            countit += 1
            mtodisplay=mtodisplay+thebuses['expected_departure_time']
            if countit != thebuslen:
                mtodisplay=mtodisplay+','
return mtodisplay

bgporter · Answer 1 · 2013-12-17T00:17:13.887

3

I'm not sure what you mean by 'resorting to lists', but something like this:

json_string = requests.get(busurl)
the_data = json_string.json()
mtodisplay= [] 

for entry in the_data['departures']:
    for thebuses in the_data['departures'][entry]:
        if thebuses['line'] == '220':
            thebuslen=len(the_data['departures'][entry])
            print 'buslen',thebuslen
            mtodisplay.append(thebuses['expected_departure_time'])

return '220 buses:\n' + ", ".join(mtodisplay)

edited Dec 17 '13 at 00:17

answered Dec 16 '13 at 22:53

bgporter

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Ah, brilliant! Thanks - exactly what I needed to learn. Much better. Will use join far more often now. Thanks @hetepeperfan too – Will Dec 16 '13 at 23:04

John La Rooy · Answer 2 · 2013-12-16T23:31:56.807

Concatenating strings like this

        mtodisplay = mtodisplay + thebuses['expected_departure_time']

Used to be very inefficient, but for a long time now, Python does reuse the string being catentated to (as long as there are no other references to it), so it's linear performance instead of the older quadratic performance which should definitely be avoided.

In this case it looks like you already have a list of items that you want to put commas between, so

','.join(some_list)

is probably more appropriate (and automatically means you don't get an extra comma at the end).

So next problem is to construct the list(could also be a generator etc.). @bgporter shows how to make the list, so I'll show the generator version

def mtodisplay(busurl):
    json_string = requests.get(busurl)
    the_data = json_string.json()
    for entry in the_data['departures']:
        for thebuses in the_data['departures'][entry]:
            if thebuses['line'] == '220':
                thebuslen=len(the_data['departures'][entry])
                print 'buslen',thebuslen
                yield thebuses['expected_departure_time']

# This is where you would normally just call the function
result = '220 buses:\n' + ','.join(mtodisplay(busurl))

Last I heard, this optimization was used on some implementations but not others. Is it now used on all implementations? Bear in mind that there are many implementations of Python, the language. — dstromberg, Dec 16 '13 at 23:31
@dstromberg, good question. It's not hard to measure by running a few benchmarks against the particular implementation — John La Rooy, Dec 16 '13 at 23:49
@gnibbler - thanks. I'm not sure about the difference between your generator and what bgporter provided. But then I don't know about generators and yields, so I'm about to [read up on them now!]((http://stackoverflow.com/questions/231767/the-python-yield-keyword-explained) — Will, Dec 17 '13 at 11:21

Concatenating strings more efficiently in Python

2 Answers2