On 09/07/2014 15:27, sssdevelop wrote:
Hello,
I have working code - but looking for better/improved code. Better coding
practices, better algorithm :)
Problem: Given sequence of increasing integers, print blocks of consecutive
integers.
Example:
Input: [10, 11, 12, 15]
Output: [10, 11, 12]
Input: [51, 53, 55, 67, 68, 91, 92, 93, 94, 99]
Outout: [67, 68], [91, 92, 93, 94]
My code looks as below:
-----------------------------
#!/usr/bin/python
a = [51, 53, 55, 67, 68, 91, 92, 93, 94, 99]
#a = []
#a = [10]
#a = [10, 11, 12, 15]
print "Input: "
print a
prev = 0
blocks = []
tmp = []
last = 0
for element in a:
if prev == 0:
prev = element
next
if element == prev + 1:
if tmp:
pass
else:
tmp.append(prev)
tmp.append(element)
else:
if tmp:
blocks.append(tmp)
tmp = []
prev = element
if tmp:
blocks.append(tmp)
if blocks:
#print "I have repeated elements and those are:"
for b in blocks:
print b
-----------------------
thank you in advance!
Adopted from here https://docs.python.org/3.0/library/itertools.html
data = [51, 53, 55, 67, 68, 91, 92, 93, 94, 99]
for k, g in groupby(enumerate(data), lambda t:t[0]-t[1]):
group = list(map(operator.itemgetter(1), g))
if len(group) > 1:
print(group)
>>>
[67, 68]
[91, 92, 93, 94]
>>>
--
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Mark Lawrence
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