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Find the most common element in a list
(27 answers)
Closed 2 years ago.
In Python, I have a list:
L = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
I want to identify the item that occurred the highest number of times. I am able to solve it but I need the fastest way to do so. I know there is a nice Pythonic answer to this.
I am surprised no-one has mentioned the simplest solution,max() with the key list.count:
max(lst,key=lst.count)
Example:
>>> lst = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
>>> max(lst,key=lst.count)
4
This works in Python 3 or 2, but note that it only returns the most frequent item and not also the frequency. Also, in the case of a draw (i.e. joint most frequent item) only a single item is returned.
Although the time complexity of using max() is worse than using Counter.most_common(1) as PM 2Ring comments, the approach benefits from a rapid C implementation and I find this approach is fastest for short lists but slower for larger ones (Python 3.6 timings shown in IPython 5.3):
In [1]: from collections import Counter
...:
...: def f1(lst):
...: return max(lst, key = lst.count)
...:
...: def f2(lst):
...: return Counter(lst).most_common(1)
...:
...: lst0 = [1,2,3,4,3]
...: lst1 = lst0[:] * 100
...:
In [2]: %timeit -n 10 f1(lst0)
10 loops, best of 3: 3.32 us per loop
In [3]: %timeit -n 10 f2(lst0)
10 loops, best of 3: 26 us per loop
In [4]: %timeit -n 10 f1(lst1)
10 loops, best of 3: 4.04 ms per loop
In [5]: %timeit -n 10 f2(lst1)
10 loops, best of 3: 75.6 us per loop
from collections import Counter
most_common,num_most_common = Counter(L).most_common(1)[0] # 4, 6 times
For older Python versions (< 2.7), you can use this recipe to create the Counter class.
In your question, you asked for the fastest way to do it. As has been demonstrated repeatedly, particularly with Python, intuition is not a reliable guide: you need to measure.
Here's a simple test of several different implementations:
import sys
from collections import Counter, defaultdict
from itertools import groupby
from operator import itemgetter
from timeit import timeit
L = [1,2,45,55,5,4,4,4,4,4,4,5456,56,6,7,67]
def max_occurrences_1a(seq=L):
"dict iteritems"
c = dict()
for item in seq:
c[item] = c.get(item, 0) + 1
return max(c.iteritems(), key=itemgetter(1))
def max_occurrences_1b(seq=L):
"dict items"
c = dict()
for item in seq:
c[item] = c.get(item, 0) + 1
return max(c.items(), key=itemgetter(1))
def max_occurrences_2(seq=L):
"defaultdict iteritems"
c = defaultdict(int)
for item in seq:
c[item] += 1
return max(c.iteritems(), key=itemgetter(1))
def max_occurrences_3a(seq=L):
"sort groupby generator expression"
return max(((k, sum(1 for i in g)) for k, g in groupby(sorted(seq))), key=itemgetter(1))
def max_occurrences_3b(seq=L):
"sort groupby list comprehension"
return max([(k, sum(1 for i in g)) for k, g in groupby(sorted(seq))], key=itemgetter(1))
def max_occurrences_4(seq=L):
"counter"
return Counter(L).most_common(1)[0]
versions = [max_occurrences_1a, max_occurrences_1b, max_occurrences_2, max_occurrences_3a, max_occurrences_3b, max_occurrences_4]
print sys.version, "\n"
for vers in versions:
print vers.__doc__, vers(), timeit(vers, number=20000)
The results on my machine:
2.7.2 (v2.7.2:8527427914a2, Jun 11 2011, 15:22:34)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)]
dict iteritems (4, 6) 0.202214956284
dict items (4, 6) 0.208412885666
defaultdict iteritems (4, 6) 0.221301078796
sort groupby generator expression (4, 6) 0.383440971375
sort groupby list comprehension (4, 6) 0.402786016464
counter (4, 6) 0.564319133759
So it appears that the Counter solution is not the fastest. And, in this case at least, groupby is faster. defaultdict is good but you pay a little bit for its convenience; it's slightly faster to use a regular dict with a get.
What happens if the list is much bigger? Adding L *= 10000 to the test above and reducing the repeat count to 200:
dict iteritems (4, 60000) 10.3451900482
dict items (4, 60000) 10.2988479137
defaultdict iteritems (4, 60000) 5.52838587761
sort groupby generator expression (4, 60000) 11.9538850784
sort groupby list comprehension (4, 60000) 12.1327362061
counter (4, 60000) 14.7495789528
Now defaultdict is the clear winner. So perhaps the cost of the 'get' method and the loss of the inplace add adds up (an examination of the generated code is left as an exercise).
But with the modified test data, the number of unique item values did not change so presumably dict and defaultdict have an advantage there over the other implementations. So what happens if we use the bigger list but substantially increase the number of unique items? Replacing the initialization of L with:
LL = [1,2,45,55,5,4,4,4,4,4,4,5456,56,6,7,67]
L = []
for i in xrange(1,10001):
L.extend(l * i for l in LL)
dict iteritems (2520, 13) 17.9935798645
dict items (2520, 13) 21.8974409103
defaultdict iteritems (2520, 13) 16.8289561272
sort groupby generator expression (2520, 13) 33.853593111
sort groupby list comprehension (2520, 13) 36.1303369999
counter (2520, 13) 22.626899004
So now Counter is clearly faster than the groupby solutions but still slower than the iteritems versions of dict and defaultdict.
The point of these examples isn't to produce an optimal solution. The point is that there often isn't one optimal general solution. Plus there are other performance criteria. The memory requirements will differ substantially among the solutions and, as the size of the input goes up, memory requirements may become the overriding factor in algorithm selection.
Bottom line: it all depends and you need to measure.
Here is a defaultdict solution that will work with Python versions 2.5 and above:
from collections import defaultdict
L = [1,2,45,55,5,4,4,4,4,4,4,5456,56,6,7,67]
d = defaultdict(int)
for i in L:
d[i] += 1
result = max(d.iteritems(), key=lambda x: x[1])
print result
# (4, 6)
# The number 4 occurs 6 times
Note if L = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 7, 7, 7, 7, 7, 56, 6, 7, 67]
then there are six 4s and six 7s. However, the result will be (4, 6) i.e. six 4s.
If you're using Python 3.8 or above, you can use either statistics.mode() to return the first mode encountered or statistics.multimode() to return all the modes.
>>> import statistics
>>> data = [1, 2, 2, 3, 3, 4]
>>> statistics.mode(data)
2
>>> statistics.multimode(data)
[2, 3]
If the list is empty, statistics.mode() throws a statistics.StatisticsError and statistics.multimode() returns an empty list.
Note before Python 3.8, statistics.mode() (introduced in 3.4) would additionally throw a statistics.StatisticsError if there is not exactly one most common value.
A simple way without any libraries or sets
def mcount(l):
n = [] #To store count of each elements
for x in l:
count = 0
for i in range(len(l)):
if x == l[i]:
count+=1
n.append(count)
a = max(n) #largest in counts list
for i in range(len(n)):
if n[i] == a:
return(l[i],a) #element,frequency
return #if something goes wrong
Perhaps the most_common() method
I obtained the best results with groupby from itertools module with this function using Python 3.5.2:
from itertools import groupby
a = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
def occurrence():
occurrence, num_times = 0, 0
for key, values in groupby(a, lambda x : x):
val = len(list(values))
if val >= occurrence:
occurrence, num_times = key, val
return occurrence, num_times
occurrence, num_times = occurrence()
print("%d occurred %d times which is the highest number of times" % (occurrence, num_times))
Output:
4 occurred 6 times which is the highest number of times
Test with timeit from timeit module.
I used this script for my test with number= 20000:
from itertools import groupby
def occurrence():
a = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
occurrence, num_times = 0, 0
for key, values in groupby(a, lambda x : x):
val = len(list(values))
if val >= occurrence:
occurrence, num_times = key, val
return occurrence, num_times
if __name__ == '__main__':
from timeit import timeit
print(timeit("occurrence()", setup = "from __main__ import occurrence", number = 20000))
Output (The best one):
0.1893607140000313
I want to throw in another solution that looks nice and is fast for short lists.
def mc(seq=L):
"max/count"
max_element = max(seq, key=seq.count)
return (max_element, seq.count(max_element))
You can benchmark this with the code provided by Ned Deily which will give you these results for the smallest test case:
3.5.2 (default, Nov 7 2016, 11:31:36)
[GCC 6.2.1 20160830]
dict iteritems (4, 6) 0.2069783889998289
dict items (4, 6) 0.20462976200065896
defaultdict iteritems (4, 6) 0.2095775119996688
sort groupby generator expression (4, 6) 0.4473949929997616
sort groupby list comprehension (4, 6) 0.4367636879997008
counter (4, 6) 0.3618192010007988
max/count (4, 6) 0.20328268999946886
But beware, it is inefficient and thus gets really slow for large lists!
Simple and best code:
def max_occ(lst,x):
count=0
for i in lst:
if (i==x):
count=count+1
return count
lst=[1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
x=max(lst,key=lst.count)
print(x,"occurs ",max_occ(lst,x),"times")
Output: 4 occurs 6 times
My (simply) code (three months studying Python):
def more_frequent_item(lst):
new_lst = []
times = 0
for item in lst:
count_num = lst.count(item)
new_lst.append(count_num)
times = max(new_lst)
key = max(lst, key=lst.count)
print("In the list: ")
print(lst)
print("The most frequent item is " + str(key) + ". Appears " + str(times) + " times in this list.")
more_frequent_item([1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67])
The output will be:
In the list:
[1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
The most frequent item is 4. Appears 6 times in this list.
if you are using numpy in your solution for faster computation use this:
import numpy as np
x = np.array([2,5,77,77,77,77,77,77,77,9,0,3,3,3,3,3])
y = np.bincount(x,minlength = max(x))
y = np.argmax(y)
print(y) #outputs 77
Following is the solution which I came up with if there are multiple characters in the string all having the highest frequency.
mystr = input("enter string: ")
#define dictionary to store characters and their frequencies
mydict = {}
#get the unique characters
unique_chars = sorted(set(mystr),key = mystr.index)
#store the characters and their respective frequencies in the dictionary
for c in unique_chars:
ctr = 0
for d in mystr:
if d != " " and d == c:
ctr = ctr + 1
mydict[c] = ctr
print(mydict)
#store the maximum frequency
max_freq = max(mydict.values())
print("the highest frequency of occurence: ",max_freq)
#print all characters with highest frequency
print("the characters are:")
for k,v in mydict.items():
if v == max_freq:
print(k)
Input: "hello people"
Output:
{'o': 2, 'p': 2, 'h': 1, ' ': 0, 'e': 3, 'l': 3}
the highest frequency of occurence: 3
the characters are:
e
l
may something like this:
testList = [1, 2, 3, 4, 2, 2, 1, 4, 4]
print(max(set(testList), key = testList.count))
Related
I have a list of caller_address elements. For each of these addresses I can get a caller_function, a function containing that caller_address. In a single function there may be more than 1 address.
So if I have a list of caller_address elements:
caller_addresses = [1, 2, 3, 4, 5, 6, 7, 8]
For each of them I can get a function:
caller_functions = [getFunctionContaining(addr) for addr in caller_addresses]
print(caller_functions)
# prints(example): ['func1', 'func1', 'func2', 'func2', 'func2', 'func2', 'func3', 'func3']
In the result I need to get a dict where keys are the functions and values are lists of addresses those functions contain. In my example in must be:
{'func1': [1, 2], 'func2': [3, 4, 5, 6], 'func3': [7, 8]}
# Means 'func1' contains addresses 1 and 2, 'func2' contains 3, 4, 5 and 6, ...
It would be great if there was a function like:
result = to_dict(lambda addr: getFunctionContaining(addr), caller_addresses)
to get the same result.
Where the first argument is the function for keys and the second argument is the list of values. Is there such function in standard library in python?
I could implement it with for loop and dict[getFunctionContaining(addr)].append(addr), but I'm looking for more pythonic way to do this.
Thanks!
Found a solution using itertools.groupby.
This solution is also faster than a solution using a loop.
import itertools
import time
def f(v):
if v < 5:
return 1
if v < 7:
return 2
return 3
def to_dict(key, list_):
out = {}
for el in list_:
out.setdefault(key(el), []).append(el)
return out
def to_dict2(key, list_):
return {k: list(v) for k, v in itertools.groupby(list_, key)}
lst = [1, 2, 3, 4, 5, 6, 7, 8] * 10**4
COUNT = 1000
def timeit(to_dict_f):
elapsed_sum = 0
for _ in range(COUNT):
elapsed_sum -= time.time()
to_dict_f(f, lst)
elapsed_sum += time.time()
return elapsed_sum / COUNT
print('Average time: ', timeit(to_dict), timeit(to_dict2))
Results:
Average time: 0.014930561065673828 0.01346096110343933
to_dict2 (itertools.groupby) on average takes less time than to_dict (loop)
I would like to know if there exists a base solution to do something like this:
for n in range(length=8, start_position= 3, direction= forward)
The problem I'm encountering is I would like the loop to continue past the final index, and pick up again at idx =0, then idx=1, etc. and stop at idx= 3, the start_position.
To give context, I seek all possible complete solutions to the n-queen problem.
Based on your latest edit, you need a "normal" range and the modulo operator:
for i in range(START, START + LEN):
do_something_with(i % LEN)
from itertools import chain
for n in chain(range(3,8), range(3)):
...
The chain() returns an iterator with 3, 4, ..., 7, 0, 1, 2
Another option for solving this is to use modular arithmetic. You could do something like this, for example:
for i in range(8)
idx = (i + 3) % 8
# use idx
This easily can be generalized to work with different lengths and offsets.
def loop_around_range(length, start_position, direction='forward'):
looped_range = [k % length for k in range(start_position, start_position+length)]
if direction == 'forward':
return looped_range
else:
return looped_range[::-1]
You could implement this for an arbitrary iterable by using itertools.cycle.
from itertools import cycle
def circular_iterator(iterable, skip=0, length=None, reverse=False):
"""Produces a full cycle of #iterable#, skipping the first #skip# elements
then tacking them on to the end.
if #iterable# does not implement #__len__#, you must provide #length#
"""
if reverse:
iterable = reversed(iterable)
cyc_iter = cycle(iterable)
for _ in range(skip):
next(cyc_iter, None)
if length:
total_length = length
else:
total_length = len(iterable)
for _ in range(total_length):
yield next(cyc_iter, None)
>>> lst = [x for x in range(1, 9)]
# [1, 2, 3, 4, 5, 6, 7, 8]
>>> list(circular_iterator(lst, skip=3))
[4, 5, 6, 7, 8, 1, 2, 3]
I have a list of integers...
[1,2,3,4,5,8,9,10,11,200,201,202]
I would like to group them into a list of lists where each sublist contains integers whose sequence has not been broken. Like this...
[[1,5],[8,11],[200,202]]
I have a rather clunky work around...
lSequenceOfNum = [1,2,3,4,5,8,9,10,11,200,201,202]
lGrouped = []
start = 0
for x in range(0,len(lSequenceOfNum)):
if x != len(lSequenceOfNum)-1:
if(lSequenceOfNum[x+1] - lSequenceOfNum[x]) > 1:
lGrouped.append([lSequenceOfNum[start],lSequenceOfNum[x]])
start = x+1
else:
lGrouped.append([lSequenceOfNum[start],lSequenceOfNum[x]])
print lGrouped
It is the best I could do. Is there a more "pythonic" way to do this? Thanks..
Assuming the list will always be in ascending order:
from itertools import groupby, count
numberlist = [1,2,3,4,5,8,9,10,11,200,201,202]
def as_range(g):
l = list(g)
return l[0], l[-1]
print [as_range(g) for _, g in groupby(numberlist, key=lambda n, c=count(): n-next(c))]
I realised I had overcomplicated this a little, far easier to just count manually than use a slightly convoluted generator:
def ranges(seq):
start, end = seq[0], seq[0]
count = start
for item in seq:
if not count == item:
yield start, end
start, end = item, item
count = item
end = item
count += 1
yield start, end
print(list(ranges([1,2,3,4,5,8,9,10,11,200,201,202])))
Producing:
[(1, 5), (8, 11), (200, 202)]
This method is pretty fast:
This method (and the old one, they perform almost exactly the same):
python -m timeit -s "from test import ranges" "ranges([1,2,3,4,5,8,9,10,11,200,201,202])"
1000000 loops, best of 3: 0.47 usec per loop
Jeff Mercado's Method:
python -m timeit -s "from test import as_range; from itertools import groupby, count" "[as_range(g) for _, g in groupby([1,2,3,4,5,8,9,10,11,200,201,202], key=lambda n, c=count(): n-next(c))]"
100000 loops, best of 3: 11.1 usec per loop
That's over 20x faster - although, naturally, unless speed matters this isn't a real concern.
My old solution using generators:
import itertools
def resetable_counter(start):
while True:
for i in itertools.count(start):
reset = yield i
if reset:
start = reset
break
def ranges(seq):
start, end = seq[0], seq[0]
counter = resetable_counter(start)
for count, item in zip(counter, seq): #In 2.x: itertools.izip(counter, seq)
if not count == item:
yield start, end
start, end = item, item
counter.send(item)
end = item
yield start, end
print(list(ranges([1,2,3,4,5,8,9,10,11,200,201,202])))
Producing:
[(1, 5), (8, 11), (200, 202)]
You can do this efficiently in three steps
given
list1=[1,2,3,4,5,8,9,10,11,200,201,202]
Calculate the discontinuity
[1,2,3,4,5,8,9,10,11 ,200,201,202]
- [1,2,3,4,5,8,9 ,10 ,11 ,200,201,202]
----------------------------------------
[1,1,1,1,3,1,1 ,1 ,189,1 ,1]
(index) 1 2 3 4 5 6 7 8 9 10 11
* *
rng = [i+1 for i,e in enumerate((x-y for x,y in zip(list1[1:],list1))) if e!=1]
>>> rng
[5, 9]
Add the boundaries
rng = [0] + rng + [len(list1)]
>>> rng
[0, 5, 9,12]
now calculate the actual continuity ranges
[(list1[i],list1[j-1]) for i,j in zip(list2,list2[1:])]
[(1, 5), (8, 11), (200, 202)]
LB [0, 5, 9, 12]
UB [0, 5, 9, 12]
-----------------------
indexes (LB,UB-1) (0,4) (5,8) (9,11)
The question is quite old, but I thought I'll share my solution anyway
Assuming import numpy as np
a = [1,2,3,4,5,8,9,10,11,200,201,202]
np.split(a, array(np.add(np.where(diff(a)>1),1)).tolist()[0])
pseudo code (with off-by-one errors to fix):
jumps = new array;
for idx from 0 to len(array)
if array[idx] != array[idx+1] then jumps.push(idx);
I think this is actually a case where it makes sense to work with the indices (as in C, before java/python/perl/etc. improved upon this) instead of the objects in the array.
Here's a version that should be easy to read:
def close_range(el, it):
while True:
el1 = next(it, None)
if el1 != el + 1:
return el, el1
el = el1
def compress_ranges(seq):
iterator = iter(seq)
left = next(iterator, None)
while left is not None:
right, left1 = close_range(left, iterator)
yield (left, right)
left = left1
list(compress_ranges([1, 2, 3, 4, 5, 8, 9, 10, 11, 200, 201, 202]))
Similar questions:
Python - find incremental numbered sequences with a list comprehension
Pythonic way to convert a list of integers into a string of comma-separated ranges
input = [1, 2, 3, 4, 8, 10, 11, 12, 17]
i, ii, result = iter(input), iter(input[1:]), [[input[0]]]
for x, y in zip(i,ii):
if y-x != 1:
result.append([y])
else:
result[-1].append(y)
>>> result
[[1, 2, 3, 4], [8], [10, 11, 12], [17]]
>>> print ", ".join("-".join(map(str,(g[0],g[-1])[:len(g)])) for g in result)
1-4, 8, 10-12, 17
>>> [(g[0],g[-1])[:len(g)] for g in result]
[(1, 4), (8,), (10, 12), (17,)]
Im trying to write a function that creates set of dynamic sublists each containing 5 elements from a list passed to it.Here's my attempt at the code
def sublists(seq):
i=0
x=[]
while i<len(seq)-1:
j=0
while j<5:
X.append(seq[i]) # How do I change X after it reaches size 5?
#return set of sublists
EDIT:
Sample input: [1,2,3,4,5,6,7,8,9,10]
Expected output: [[1,2,3,4,5],[6,7,8,9,10]]
Well, for starters, you'll need to (or at least should) have two lists, a temporary one and a permanent one that you return (Also you will need to increase j and i or, more practically, use a for loop, but I assume you just forgot to post that).
EDIT removed first code as the style given doesn't match easily with the expected results, see other two possibilities.
Or, more sensibly:
def sublists(seq):
x=[]
for i in range(0,len(seq),5):
x.append(seq[i:i+5])
return x
Or, more sensibly again, a simple list comprehension:
def sublists(seq):
return [seq[i:i+5] for i in range(0,len(seq),5)]
When given the list:
l = [1,2,3,4,5,6,7,8,9,10]
They will return
[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
Have you considered using itertools.combinations(...)?
For example:
>>> from itertools import combinations
>>> l = [1,2,3,4,5,6]
>>> list(combinations(l, 5))
[(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6), (1, 3, 4, 5, 6), (2, 3, 4, 5, 6)]
By "dynamic sublists", do you mean break up the list into groups of five elements? This is similar to your approach:
def sublists(lst, n):
ret = []
i = 0
while i < len(lst):
ret.append(seq[i:i+n])
i += n
return ret
Or, using iterators:
def sublists(seq, n):
it = iter(seq)
while True:
r = list(itertools.islice(it, 5))
if not r:
break
yield r
which will return an iterator of lists over list of length up to five. (If you took out the list call, weird things would happen if you didn't access the iterators in the same order.)
I have a sumranges() function, which sums all the ranges of consecutive numbers found in a tuple of tuples. To illustrate:
def sumranges(nums):
return sum([sum([1 for j in range(len(nums[i])) if
nums[i][j] == 0 or
nums[i][j - 1] + 1 != nums[i][j]]) for
i in range(len(nums))])
>>> nums = ((1, 2, 3, 4), (1, 5, 6), (19, 20, 24, 29, 400))
>>> print sumranges(nums)
7
As you can see, it returns the number of ranges of consecutive digits within the tuple, that is: len((1, 2, 3, 4), (1), (5, 6), (19, 20), (24), (29), (400)) = 7. The tuples are always ordered.
My problem is that my sumranges() is terrible. I hate looking at it. I'm currently just iterating through the tuple and each subtuple, assigning a 1 if the number is not (1 + previous number), and summing the total. I feel like I am missing a much easier way to accomplish my stated objective. Does anyone know a more pythonic way to do this?
Edit: I have benchmarked all the answers given thus far. Thanks to all of you for your answers.
The benchmarking code is as follows, using a sample size of 100K:
from time import time
from random import randrange
nums = [sorted(list(set(randrange(1, 10) for i in range(10)))) for
j in range(100000)]
for func in sumranges, alex, matt, redglyph, ephemient, ferdinand:
start = time()
result = func(nums)
end = time()
print ', '.join([func.__name__, str(result), str(end - start) + ' s'])
Results are as follows. Actual answer shown to verify that all functions return the correct answer:
sumranges, 250281, 0.54171204567 s
alex, 250281, 0.531121015549 s
matt, 250281, 0.843333005905 s
redglyph, 250281, 0.366822004318 s
ephemient, 250281, 0.805964946747 s
ferdinand, 250281, 0.405596971512 s
RedGlyph does edge out in terms of speed, but the simplest answer is probably Ferdinand's, and probably wins for most pythonic.
My 2 cents:
>>> sum(len(set(x - i for i, x in enumerate(t))) for t in nums)
7
It's basically the same idea as descriped in Alex' post, but using a set instead of itertools.groupby, resulting in a shorter expression. Since sets are implemented in C and len() of a set runs in constant time, this should also be pretty fast.
Consider:
>>> nums = ((1, 2, 3, 4), (1, 5, 6), (19, 20, 24, 29, 400))
>>> flat = [[(x - i) for i, x in enumerate(tu)] for tu in nums]
>>> print flat
[[1, 1, 1, 1], [1, 4, 4], [19, 19, 22, 26, 396]]
>>> import itertools
>>> print sum(1 for tu in flat for _ in itertools.groupby(tu))
7
>>>
we "flatten" the "increasing ramps" of interest by subtracting the index from the value, turning them into consecutive "runs" of identical values; then we identify and could the "runs" with the precious itertools.groupby. This seems to be a pretty elegant (and speedy) solution to your problem.
Just to show something closer to your original code:
def sumranges(nums):
return sum( (1 for i in nums
for j, v in enumerate(i)
if j == 0 or v != i[j-1] + 1) )
The idea here was to:
avoid building intermediate lists but use a generator instead, it will save some resources
avoid using indices when you already have selected a subelement (i and v above).
The remaining sum() is still necessary with my example though.
Here's my attempt:
def ranges(ls):
for l in ls:
consec = False
for (a,b) in zip(l, l[1:]+(None,)):
if b == a+1:
consec = True
if b is not None and b != a+1:
consec = False
if consec:
yield 1
'''
>>> nums = ((1, 2, 3, 4), (1, 5, 6), (19, 20, 24, 29, 400))
>>> print sum(ranges(nums))
7
'''
It looks at the numbers pairwise, checking if they are a consecutive pair (unless it's at the last element of the list). Each time there's a consecutive pair of numbers it yields 1.
This could probably be put together in a more compact form, but I think clarity would suffer:
def pairs(seq):
for i in range(1,len(seq)):
yield (seq[i-1], seq[i])
def isadjacent(pair):
return pair[0]+1 == pair[1]
def sumrange(seq):
return 1 + sum([1 for pair in pairs(seq) if not isadjacent(pair)])
def sumranges(nums):
return sum([sumrange(seq) for seq in nums])
nums = ((1, 2, 3, 4), (1, 5, 6), (19, 20, 24, 29, 400))
print sumranges(nums) # prints 7
You could probably do this better if you had an IntervalSet class because then you would scan through your ranges to build your IntervalSet, then just use the count of set members.
Some tasks don't always lend themselves to neat code, particularly if you need to write the code for performance.
There is a formula for this, the sum of the first n numbers, 1+ 2+ ... + n = n(n+1) / 2 . Then if you want to have the sum of i-j then it is (j(j+1)/2) - (i(i+1)/2) this I am sure simplifies but you can work that out. It might not be pythonic but it is what I would use.