After iterating through a list in Python, print out remainder

Question

I have a list of numbers that is subject to change. I would like to cut it up into chunks of 16, which is what the below accomplishes, but unless exactly divisible by 16 (not often), I get leftovers. I'd like to know how to output the remaining items after the bulk has been processed.

s = ["002","003","005","006","007","008","009","010","012","013","014","015","016","017","018","019","023","024","025","026","027","028","029","031","032","033","034","035","037","038","040","041","042","043","044","045","046","047","048","049","050","051","052","053","055","059","060","063","064","065","066","067","069","070","071","072","073","074","075","076","077","078","079","080","081","082","083","086","088","089","090","091","092","093","094","095","096","097","098","099","100","101","102","105","106","107","109","110","111","112","113","114","115","117","118","120","122","123","124","125","127","128","129","130","131","133","134","135","136","137","138","139","140","143","145","147","148","149","150","152","153","154","155","157","158","160","161","163","165","166","167","169","170","172","173","174","175","177","178","179","181","182","183","185","186","187","188","189","191","192","193","194","195","196","201","202","203","204","205","206","207","208","210","211","212","213","214","215","216","217","218","220","221","222","223","225","226","227","228","229","230","231","232","233","234","235","236","237","238","240","241","242","243","244","245","247","248","300","302"]

count = 0
store_count = []
for i in s:
    count += 1
    store_count.append(i)
    if count % 16 == 0 :
        print(store_count)
        store_count = []

Output appears like so:

['207', '208', '210', '211', '212', '213', '214', '215', '216', '217', '218', '220', '221', '222', '223', '225']
['226', '227', '228', '229', '230', '231', '232', '233', '234', '235', '236', '237', '238', '240', '241', '242']

Much easier (and performant) to use slicing than incrementing, e.g. `for x in (s[i:i+16] for i in range(0, len(s), 16)): print(x)` — AChampion, Oct 17 '15 at 05:37
@AChampion Excellent point. Feel free to put it up as an answer. — David Metcalfe, Oct 17 '15 at 05:42

score 2 · Accepted Answer · answered Oct 17 '15 at 05:25

After the loop, check if the store_count has items:

for i in s:
    count += 1
    store_count.append(i)
    if count % 16 == 0 :
        print(store_count)
        store_count = []
if store_count:       # <---- Check leftover
    print(store_count) # <----

BTW, instead of increasing the count yourself, you can use enumerate:

store_count = []
for count, i in enumerate(s, 1):  # <----  `count` will be 1, 2, 3, ...
    store_count.append(i)
    if count % 16 == 0 :
        print(store_count)
        store_count = []
if store_count:
    print(store_count)

Great extra tip on enumerate. Did not know that was a thing. — David Metcalfe, Oct 17 '15 at 05:36

ZdaR · Answer 2 · 2015-10-17T05:26:26.407

0

You may simply get the leftover values as :

if len(s)%16:
    leftover = s[-(max(len(s)%16, 1)):]

edited Oct 17 '15 at 05:26

answered Oct 17 '15 at 05:23

ZdaR

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After iterating through a list in Python, print out remainder

2 Answers2