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I want to declare a set of variables with PuLP which contains all the possible combinations of the following lists:
month = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
manufacturer = ['China', 'Mexico', 'Taiwan']
demand = ['London', 'Paris', 'Milan']
Then, I will have a dictionary (for example) as follow:
'1.China.London', '1.China.Paris',...
I tried with the following code, but I don't know how to store all the combinations.
vlbs = {}
for key in month:
for kay in manufacturer:
for eyk in demand:
vlbs = (str(key)+'.'+str(kay)+'.'+str(eyk))
First, I'm not getting properly the dictionary vlbs. And later on:
variables = {var: pl.LpVariable(var, lowBound = 0) for var in vlbs}
How can I solve it properly??
You can use a tuple as a dictionary key. I think this makes filtering/searching easier than doing a bunch of string concatenating and splitting.
months = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
manufacturers = ['China', 'Mexico', 'Taiwan']
demands = ['London', 'Paris', 'Milan']
var_dict = {}
for month in months:
for manufacturer in manufacturers:
for demand in demands:
combo = (month, manufacturer, demand)
var_name = '.'.join([str(c) for c in combo])
var_dict[combo] = LpVariable(var_name, lowBound=0)
# add constraint for month 1
model += lpSum([k for k in var_dict if k[0] == 1]) <= 50
I'd like to identify groups of consecutive numbers in a list, so that:
myfunc([2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20])
Returns:
[(2,5), (12,17), 20]
And was wondering what the best way to do this was (particularly if there's something inbuilt into Python).
Edit: Note I originally forgot to mention that individual numbers should be returned as individual numbers, not ranges.
EDIT 2: To answer the OP new requirement
ranges = []
for key, group in groupby(enumerate(data), lambda (index, item): index - item):
group = map(itemgetter(1), group)
if len(group) > 1:
ranges.append(xrange(group[0], group[-1]))
else:
ranges.append(group[0])
Output:
[xrange(2, 5), xrange(12, 17), 20]
You can replace xrange with range or any other custom class.
Python docs have a very neat recipe for this:
from operator import itemgetter
from itertools import groupby
data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
print(map(itemgetter(1), g))
Output:
[2, 3, 4, 5]
[12, 13, 14, 15, 16, 17]
If you want to get the exact same output, you can do this:
ranges = []
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
group = map(itemgetter(1), g)
ranges.append((group[0], group[-1]))
output:
[(2, 5), (12, 17)]
EDIT: The example is already explained in the documentation but maybe I should explain it more:
The key to the solution is
differencing with a range so that
consecutive numbers all appear in same
group.
If the data was: [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
Then groupby(enumerate(data), lambda (i,x):i-x) is equivalent of the following:
groupby(
[(0, 2), (1, 3), (2, 4), (3, 5), (4, 12),
(5, 13), (6, 14), (7, 15), (8, 16), (9, 17)],
lambda (i,x):i-x
)
The lambda function subtracts the element index from the element value. So when you apply the lambda on each item. You'll get the following keys for groupby:
[-2, -2, -2, -2, -8, -8, -8, -8, -8, -8]
groupby groups elements by equal key value, so the first 4 elements will be grouped together and so forth.
I hope this makes it more readable.
python 3 version may be helpful for beginners
import the libraries required first
from itertools import groupby
from operator import itemgetter
ranges =[]
for k,g in groupby(enumerate(data),lambda x:x[0]-x[1]):
group = (map(itemgetter(1),g))
group = list(map(int,group))
ranges.append((group[0],group[-1]))
more_itertools.consecutive_groups was added in version 4.0.
Demo
import more_itertools as mit
iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
[list(group) for group in mit.consecutive_groups(iterable)]
# [[2, 3, 4, 5], [12, 13, 14, 15, 16, 17], [20]]
Code
Applying this tool, we make a generator function that finds ranges of consecutive numbers.
def find_ranges(iterable):
"""Yield range of consecutive numbers."""
for group in mit.consecutive_groups(iterable):
group = list(group)
if len(group) == 1:
yield group[0]
else:
yield group[0], group[-1]
iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
list(find_ranges(iterable))
# [(2, 5), (12, 17), 20]
The source implementation emulates a classic recipe (as demonstrated by #Nadia Alramli).
Note: more_itertools is a third-party package installable via pip install more_itertools.
The "naive" solution which I find somewhat readable atleast.
x = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 22, 25, 26, 28, 51, 52, 57]
def group(L):
first = last = L[0]
for n in L[1:]:
if n - 1 == last: # Part of the group, bump the end
last = n
else: # Not part of the group, yield current group and start a new
yield first, last
first = last = n
yield first, last # Yield the last group
>>>print list(group(x))
[(2, 5), (12, 17), (22, 22), (25, 26), (28, 28), (51, 52), (57, 57)]
Assuming your list is sorted:
>>> from itertools import groupby
>>> def ranges(lst):
pos = (j - i for i, j in enumerate(lst))
t = 0
for i, els in groupby(pos):
l = len(list(els))
el = lst[t]
t += l
yield range(el, el+l)
>>> lst = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
>>> list(ranges(lst))
[range(2, 6), range(12, 18)]
Here it is something that should work, without any import needed:
def myfunc(lst):
ret = []
a = b = lst[0] # a and b are range's bounds
for el in lst[1:]:
if el == b+1:
b = el # range grows
else: # range ended
ret.append(a if a==b else (a,b)) # is a single or a range?
a = b = el # let's start again with a single
ret.append(a if a==b else (a,b)) # corner case for last single/range
return ret
Please note that the code using groupby doesn't work as given in Python 3 so use this.
for k, g in groupby(enumerate(data), lambda x:x[0]-x[1]):
group = list(map(itemgetter(1), g))
ranges.append((group[0], group[-1]))
This doesn't use a standard function - it just iiterates over the input, but it should work:
def myfunc(l):
r = []
p = q = None
for x in l + [-1]:
if x - 1 == q:
q += 1
else:
if p:
if q > p:
r.append('%s-%s' % (p, q))
else:
r.append(str(p))
p = q = x
return '(%s)' % ', '.join(r)
Note that it requires that the input contains only positive numbers in ascending order. You should validate the input, but this code is omitted for clarity.
import numpy as np
myarray = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
sequences = np.split(myarray, np.array(np.where(np.diff(myarray) > 1)[0]) + 1)
l = []
for s in sequences:
if len(s) > 1:
l.append((np.min(s), np.max(s)))
else:
l.append(s[0])
print(l)
Output:
[(2, 5), (12, 17), 20]
I think this way is simpler than any of the answers I've seen here (Edit: fixed based on comment from Pleastry):
data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
starts = [x for x in data if x-1 not in data and x+1 in data]
ends = [x for x in data if x-1 in data and x+1 not in data and x not in starts]
singles = [x for x in data if x-1 not in data and x+1 not in data]
list(zip(starts, ends)) + singles
Output:
[(2, 5), (12, 17), 20]
Edited:
As #dawg notes, this is O(n**2). One option to improve performance would be to convert the original list to a set (and also the starts list to a set) i.e.
data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
data_as_set = set(data)
starts = [x for x in data_as_set if x-1 not in data_as_set and x+1 in data_as_set]
startset = set(starts)
ends = [x for x in data_as_set if x-1 in data_as_set and x+1 not in data_as_set and x not in startset]
singles = [x for x in data_as_set if x-1 not in data_as_set and x+1 not in data_as_set]
print(list(zip(starts, ends)) + singles)
Using groupby and count from itertools gives us a short solution. The idea is that, in an increasing sequence, the difference between the index and the value will remain the same.
In order to keep track of the index, we can use an itertools.count, which makes the code cleaner as using enumerate:
from itertools import groupby, count
def intervals(data):
out = []
counter = count()
for key, group in groupby(data, key = lambda x: x-next(counter)):
block = list(group)
out.append([block[0], block[-1]])
return out
Some sample output:
print(intervals([0, 1, 3, 4, 6]))
# [[0, 1], [3, 4], [6, 6]]
print(intervals([2, 3, 4, 5]))
# [[2, 5]]
This is my method in which I tried to prioritize readability. Note that it returns a tuple of the same values if there is only one value in a group. That can be fixed easily in the second snippet I'll post.
def group(values):
"""return the first and last value of each continuous set in a list of sorted values"""
values = sorted(values)
first = last = values[0]
for index in values[1:]:
if index - last > 1: # triggered if in a new group
yield first, last
first = index # update first only if in a new group
last = index # update last on every iteration
yield first, last # this is needed to yield the last set of numbers
Here is the result of a test:
values = [0, 5, 6, 7, 12, 13, 21, 22, 23, 24, 25, 26, 30, 44, 45, 50]
result = list(group(values))
print(result)
result = [(0, 0), (5, 7), (12, 13), (21, 26), (30, 30), (44, 45), (50, 50)]
If you want to return only a single value in the case of a single value in a group, just add a conditional check to the yields:
def group(values):
"""return the first and last value of each continuous set in a list of sorted values"""
values = sorted(values)
first = last = values[0]
for index in values[1:]:
if index - last > 1: # triggered if in a new group
if first == last:
yield first
else:
yield first, last
first = index # update first only if in a new group
last = index # update last on every iteration
if first == last:
yield first
else:
yield first, last
result = [0, (5, 7), (12, 13), (21, 26), 30, (44, 45), 50]
Here's the answer I came up with. I'm writing the code for other people to understand, so I'm fairly verbose with variable names and comments.
First a quick helper function:
def getpreviousitem(mylist,myitem):
'''Given a list and an item, return previous item in list'''
for position, item in enumerate(mylist):
if item == myitem:
# First item has no previous item
if position == 0:
return None
# Return previous item
return mylist[position-1]
And then the actual code:
def getranges(cpulist):
'''Given a sorted list of numbers, return a list of ranges'''
rangelist = []
inrange = False
for item in cpulist:
previousitem = getpreviousitem(cpulist,item)
if previousitem == item - 1:
# We're in a range
if inrange == True:
# It's an existing range - change the end to the current item
newrange[1] = item
else:
# We've found a new range.
newrange = [item-1,item]
# Update to show we are now in a range
inrange = True
else:
# We were in a range but now it just ended
if inrange == True:
# Save the old range
rangelist.append(newrange)
# Update to show we're no longer in a range
inrange = False
# Add the final range found to our list
if inrange == True:
rangelist.append(newrange)
return rangelist
Example run:
getranges([2, 3, 4, 5, 12, 13, 14, 15, 16, 17])
returns:
[[2, 5], [12, 17]]
Using numpy + comprehension lists:
With numpy diff function, consequent input vector entries that their difference is not equal to one can be identified. The start and end of the input vector need to be considered.
import numpy as np
data = np.array([2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20])
d = [i for i, df in enumerate(np.diff(data)) if df!= 1]
d = np.hstack([-1, d, len(data)-1]) # add first and last elements
d = np.vstack([d[:-1]+1, d[1:]]).T
print(data[d])
Output:
[[ 2 5]
[12 17]
[20 20]]
Note: The request that individual numbers should be treated differently, (returned as individual, not ranges) was omitted. This can be reached by further post-processing the results. Usually this will make things more complex without gaining any benefit.
One-liner in Python 2.7 if interested:
x = [2, 3, 6, 7, 8, 14, 15, 19, 20, 21]
d = iter(x[:1] + sum(([i1, i2] for i1, i2 in zip(x, x[1:] + x[:1]) if i2 != i1+1), []))
print zip(d, d)
>>> [(2, 3), (6, 8), (14, 15), (19, 21)]
A short solution that works without additional imports. It accepts any iterable, sorts unsorted inputs, and removes duplicate items:
def ranges(nums):
nums = sorted(set(nums))
gaps = [[s, e] for s, e in zip(nums, nums[1:]) if s+1 < e]
edges = iter(nums[:1] + sum(gaps, []) + nums[-1:])
return list(zip(edges, edges))
Example:
>>> ranges([2, 3, 4, 7, 8, 9, 15])
[(2, 4), (7, 9), (15, 15)]
>>> ranges([-1, 0, 1, 2, 3, 12, 13, 15, 100])
[(-1, 3), (12, 13), (15, 15), (100, 100)]
>>> ranges(range(100))
[(0, 99)]
>>> ranges([0])
[(0, 0)]
>>> ranges([])
[]
This is the same as #dansalmo's solution which I found amazing, albeit a bit hard to read and apply (as it's not given as a function).
Note that it could easily be modified to spit out "traditional" open ranges [start, end), by e.g. altering the return statement:
return [(s, e+1) for s, e in zip(edges, edges)]
I copied this answer over from another question that was marked as a duplicate of this one with the intention to make it easier findable (after I just now searched again for this topic, finding only the question here at first and not being satisfied with the answers given).
The versions by Mark Byers, Andrea Ambu, SilentGhost, Nadia Alramli, and truppo are simple and fast. The 'truppo' version encouraged me to write a version that retains the same nimble behavior while handling step sizes other than 1 (and lists as singletons elements that don't extend more than 1 step with a given step size). It is given here.
>>> list(ranges([1,2,3,4,3,2,1,3,5,7,11,1,2,3]))
[(1, 4, 1), (3, 1, -1), (3, 7, 2), 11, (1, 3, 1)]
Not the best approach , but here is my 2 cents
def getConsecutiveValues2(arr):
x = ""
final = []
end = 0
start = 0
for i in range(1,len(arr)) :
if arr[i] - arr[i-1] == 1 :
end = i
else :
print(start,end)
final.append(arr[start:end+1])
start = i
if i == len(arr) - 1 :
final.append(arr[start:end+1])
return final
x = [1,2,3,5,6,8,9,10,11,12]
print(getConsecutiveValues2(x))
>> [[1, 2, 3], [5, 6], [8, 9, 10, 11]]
This implementation works for regular or irregular steps
I needed to achieve the same thing but with the slight difference where steps can be irregular. this is my implementation
def ranges(l):
if not len(l):
return range(0,0)
elif len(l)==1:
return range(l[0],l[0]+1)
# get steps
sl = sorted(l)
steps = [i-j for i,j in zip(sl[1:],sl[:-1])]
# get unique steps indexes range
groups = [[0,0,steps[0]],]
for i,s in enumerate(steps):
if s==groups[-1][-1]:
groups[-1][1] = i+1
else:
groups.append( [i+1,i+1,s] )
g2 = groups[-2]
if g2[0]==g2[1]:
if sl[i+1]-sl[i]==s:
_=groups.pop(-2)
groups[-1][0] = i
# create list of ranges
return [range(sl[i],sl[j]+s,s) if s!=0 else [sl[i]]*(j+1-i) for i,j,s in groups]
Here's an example
from timeit import timeit
# for regular ranges
l = list(range(1000000))
ranges(l)
>>> [range(0, 1000000)]
l = list(range(10)) + list(range(20,25)) + [1,2,3]
ranges(l)
>>> [range(0, 2), range(1, 3), range(2, 4), range(3, 10), range(20, 25)]
sorted(l);[list(i) for i in ranges(l)]
>>> [0, 1, 1, 2, 2, 3, 3, 4, 5, 6, 7, 8, 9, 20, 21, 22, 23, 24]
>>> [[0, 1], [1, 2], [2, 3], [3, 4, 5, 6, 7, 8, 9], [20, 21, 22, 23, 24]]
# for irregular steps list
l = [1, 3, 5, 7, 10, 11, 12, 100, 200, 300, 400, 60, 99, 4000,4001]
ranges(l)
>>> [range(1, 9, 2), range(10, 13), range(60, 138, 39), range(100, 500, 100), range(4000, 4002)]
## Speed test
timeit("ranges(l)","from __main__ import ranges,l", number=1000)/1000
>>> 9.303160999934334e-06
Yet another solution if you expect your input to be a set:
def group_years(years):
consecutive_years = []
for year in years:
close = {y for y in years if abs(y - year) == 1}
for group in consecutive_years:
if len(close.intersection(group)):
group |= close
break
else:
consecutive_years.append({year, *close})
return consecutive_years
Example:
group_years({2016, 2017, 2019, 2020, 2022})
Out[54]: [{2016, 2017}, {2019, 2020}, {2022}]
I have a list consist of tuple items like (id, cost, clicks, views) as below:
statistic_data_list = [(12324, 9, 6, 9), (12325, 11, 5, 3), (12326, 10, 7, 2)]
And I want to get the item's id which meet following conditions:
1 when not all cost of item is equal to 0, get the item's id which cost is lowest.
2 when all cost of item is equal 0, then if not all clicks of item is equal to 0, get the item's id which clicks is lowest.
3 when all clicks of item is equal to 0, then if not all views of item is equal to 0, get the item's id which views is lowest.
# (1)
# input:
[(12324, 9, 6, 9), (12325, 11, 5, 3), (12326, 10, 7, 2)]
# expected result:
12324 # (whose cost is lowest)
# (2)
# input:
[(12324, 0, 6, 9), (12325, 0, 5, 3), (12326, 0, 7, 2)]
# expected result:
12325 # (whose clicks is lowest when all cost is 0)
# (3)
# input:
[(12324, 0, 0, 9), (12325, 0, 0, 3), (12326, 0, 0, 2)]
# expected result:
12326 # (whose views is lowest when all cost is 0 also clicks)
How can I get the specified item's id more efficiently?
# My attemp so far
cost_clicks_views_list = [(12324, 9, 6, 9), (12325, 11, 5, 3), (12326, 10, 7, 2)]
len_cost_not_0 = len(list(filter(lambda item: item[1], cost_clicks_views_list)))
len_clicks_not_0 = len(list(filter(lambda item: item[2], cost_clicks_views_list)))
len_views_not_0 = len(list(filter(lambda item: item[3], cost_clicks_views_list)))
if len_cost_not_0:
min_cost_id_list = [ item[0] for item in cost_clicks_views_list if item[1]==min([i[1] for i in cost_clicks_views_list]) ]
print(min_cost_id_list) # [(12324]
else:
if len_clicks_not_0:
min_clicks_id_list = [item[0] for item in cost_clicks_views_list if item[2] == min([i[2] for i in cost_clicks_views_list])]
print(min_clicks_id_list) # [(12325]
else:
if len_views_not_0:
min_views_id_list = [item[0] for item in cost_clicks_views_list if item[3] == min([i[3] for i in cost_clicks_views_list])]
print(min_views_id_list) # [12326]
Any commentary is very welcome. great thanks.
You can use list comprehension for checking these. When at least one cost among all items is not equal to zero, to get the item's id whose cost is lowest, try this :
sdl = [(12324, 10, 0.6, 9), (12325, 11, 0.5, 3), (12326, 10, 0.7, 2)]
a = [j[0] for j in sdl if j[1]==min([k[1] for k in sdl if all([True if i[1] != 0 else False for i in sdl])])]
OUTPUT :
a = [12324, 12326]
Here all three items have non-zero cost and lowest cost is 10 corresponding to which there are two ids 12324 and 12326.
I'd like to identify groups of consecutive numbers in a list, so that:
myfunc([2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20])
Returns:
[(2,5), (12,17), 20]
And was wondering what the best way to do this was (particularly if there's something inbuilt into Python).
Edit: Note I originally forgot to mention that individual numbers should be returned as individual numbers, not ranges.
EDIT 2: To answer the OP new requirement
ranges = []
for key, group in groupby(enumerate(data), lambda (index, item): index - item):
group = map(itemgetter(1), group)
if len(group) > 1:
ranges.append(xrange(group[0], group[-1]))
else:
ranges.append(group[0])
Output:
[xrange(2, 5), xrange(12, 17), 20]
You can replace xrange with range or any other custom class.
Python docs have a very neat recipe for this:
from operator import itemgetter
from itertools import groupby
data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
print(map(itemgetter(1), g))
Output:
[2, 3, 4, 5]
[12, 13, 14, 15, 16, 17]
If you want to get the exact same output, you can do this:
ranges = []
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
group = map(itemgetter(1), g)
ranges.append((group[0], group[-1]))
output:
[(2, 5), (12, 17)]
EDIT: The example is already explained in the documentation but maybe I should explain it more:
The key to the solution is
differencing with a range so that
consecutive numbers all appear in same
group.
If the data was: [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
Then groupby(enumerate(data), lambda (i,x):i-x) is equivalent of the following:
groupby(
[(0, 2), (1, 3), (2, 4), (3, 5), (4, 12),
(5, 13), (6, 14), (7, 15), (8, 16), (9, 17)],
lambda (i,x):i-x
)
The lambda function subtracts the element index from the element value. So when you apply the lambda on each item. You'll get the following keys for groupby:
[-2, -2, -2, -2, -8, -8, -8, -8, -8, -8]
groupby groups elements by equal key value, so the first 4 elements will be grouped together and so forth.
I hope this makes it more readable.
python 3 version may be helpful for beginners
import the libraries required first
from itertools import groupby
from operator import itemgetter
ranges =[]
for k,g in groupby(enumerate(data),lambda x:x[0]-x[1]):
group = (map(itemgetter(1),g))
group = list(map(int,group))
ranges.append((group[0],group[-1]))
more_itertools.consecutive_groups was added in version 4.0.
Demo
import more_itertools as mit
iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
[list(group) for group in mit.consecutive_groups(iterable)]
# [[2, 3, 4, 5], [12, 13, 14, 15, 16, 17], [20]]
Code
Applying this tool, we make a generator function that finds ranges of consecutive numbers.
def find_ranges(iterable):
"""Yield range of consecutive numbers."""
for group in mit.consecutive_groups(iterable):
group = list(group)
if len(group) == 1:
yield group[0]
else:
yield group[0], group[-1]
iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
list(find_ranges(iterable))
# [(2, 5), (12, 17), 20]
The source implementation emulates a classic recipe (as demonstrated by #Nadia Alramli).
Note: more_itertools is a third-party package installable via pip install more_itertools.
The "naive" solution which I find somewhat readable atleast.
x = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 22, 25, 26, 28, 51, 52, 57]
def group(L):
first = last = L[0]
for n in L[1:]:
if n - 1 == last: # Part of the group, bump the end
last = n
else: # Not part of the group, yield current group and start a new
yield first, last
first = last = n
yield first, last # Yield the last group
>>>print list(group(x))
[(2, 5), (12, 17), (22, 22), (25, 26), (28, 28), (51, 52), (57, 57)]
Assuming your list is sorted:
>>> from itertools import groupby
>>> def ranges(lst):
pos = (j - i for i, j in enumerate(lst))
t = 0
for i, els in groupby(pos):
l = len(list(els))
el = lst[t]
t += l
yield range(el, el+l)
>>> lst = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
>>> list(ranges(lst))
[range(2, 6), range(12, 18)]
Here it is something that should work, without any import needed:
def myfunc(lst):
ret = []
a = b = lst[0] # a and b are range's bounds
for el in lst[1:]:
if el == b+1:
b = el # range grows
else: # range ended
ret.append(a if a==b else (a,b)) # is a single or a range?
a = b = el # let's start again with a single
ret.append(a if a==b else (a,b)) # corner case for last single/range
return ret
Please note that the code using groupby doesn't work as given in Python 3 so use this.
for k, g in groupby(enumerate(data), lambda x:x[0]-x[1]):
group = list(map(itemgetter(1), g))
ranges.append((group[0], group[-1]))
This doesn't use a standard function - it just iiterates over the input, but it should work:
def myfunc(l):
r = []
p = q = None
for x in l + [-1]:
if x - 1 == q:
q += 1
else:
if p:
if q > p:
r.append('%s-%s' % (p, q))
else:
r.append(str(p))
p = q = x
return '(%s)' % ', '.join(r)
Note that it requires that the input contains only positive numbers in ascending order. You should validate the input, but this code is omitted for clarity.
import numpy as np
myarray = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
sequences = np.split(myarray, np.array(np.where(np.diff(myarray) > 1)[0]) + 1)
l = []
for s in sequences:
if len(s) > 1:
l.append((np.min(s), np.max(s)))
else:
l.append(s[0])
print(l)
Output:
[(2, 5), (12, 17), 20]
I think this way is simpler than any of the answers I've seen here (Edit: fixed based on comment from Pleastry):
data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
starts = [x for x in data if x-1 not in data and x+1 in data]
ends = [x for x in data if x-1 in data and x+1 not in data and x not in starts]
singles = [x for x in data if x-1 not in data and x+1 not in data]
list(zip(starts, ends)) + singles
Output:
[(2, 5), (12, 17), 20]
Edited:
As #dawg notes, this is O(n**2). One option to improve performance would be to convert the original list to a set (and also the starts list to a set) i.e.
data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
data_as_set = set(data)
starts = [x for x in data_as_set if x-1 not in data_as_set and x+1 in data_as_set]
startset = set(starts)
ends = [x for x in data_as_set if x-1 in data_as_set and x+1 not in data_as_set and x not in startset]
singles = [x for x in data_as_set if x-1 not in data_as_set and x+1 not in data_as_set]
print(list(zip(starts, ends)) + singles)
Using groupby and count from itertools gives us a short solution. The idea is that, in an increasing sequence, the difference between the index and the value will remain the same.
In order to keep track of the index, we can use an itertools.count, which makes the code cleaner as using enumerate:
from itertools import groupby, count
def intervals(data):
out = []
counter = count()
for key, group in groupby(data, key = lambda x: x-next(counter)):
block = list(group)
out.append([block[0], block[-1]])
return out
Some sample output:
print(intervals([0, 1, 3, 4, 6]))
# [[0, 1], [3, 4], [6, 6]]
print(intervals([2, 3, 4, 5]))
# [[2, 5]]
This is my method in which I tried to prioritize readability. Note that it returns a tuple of the same values if there is only one value in a group. That can be fixed easily in the second snippet I'll post.
def group(values):
"""return the first and last value of each continuous set in a list of sorted values"""
values = sorted(values)
first = last = values[0]
for index in values[1:]:
if index - last > 1: # triggered if in a new group
yield first, last
first = index # update first only if in a new group
last = index # update last on every iteration
yield first, last # this is needed to yield the last set of numbers
Here is the result of a test:
values = [0, 5, 6, 7, 12, 13, 21, 22, 23, 24, 25, 26, 30, 44, 45, 50]
result = list(group(values))
print(result)
result = [(0, 0), (5, 7), (12, 13), (21, 26), (30, 30), (44, 45), (50, 50)]
If you want to return only a single value in the case of a single value in a group, just add a conditional check to the yields:
def group(values):
"""return the first and last value of each continuous set in a list of sorted values"""
values = sorted(values)
first = last = values[0]
for index in values[1:]:
if index - last > 1: # triggered if in a new group
if first == last:
yield first
else:
yield first, last
first = index # update first only if in a new group
last = index # update last on every iteration
if first == last:
yield first
else:
yield first, last
result = [0, (5, 7), (12, 13), (21, 26), 30, (44, 45), 50]
Here's the answer I came up with. I'm writing the code for other people to understand, so I'm fairly verbose with variable names and comments.
First a quick helper function:
def getpreviousitem(mylist,myitem):
'''Given a list and an item, return previous item in list'''
for position, item in enumerate(mylist):
if item == myitem:
# First item has no previous item
if position == 0:
return None
# Return previous item
return mylist[position-1]
And then the actual code:
def getranges(cpulist):
'''Given a sorted list of numbers, return a list of ranges'''
rangelist = []
inrange = False
for item in cpulist:
previousitem = getpreviousitem(cpulist,item)
if previousitem == item - 1:
# We're in a range
if inrange == True:
# It's an existing range - change the end to the current item
newrange[1] = item
else:
# We've found a new range.
newrange = [item-1,item]
# Update to show we are now in a range
inrange = True
else:
# We were in a range but now it just ended
if inrange == True:
# Save the old range
rangelist.append(newrange)
# Update to show we're no longer in a range
inrange = False
# Add the final range found to our list
if inrange == True:
rangelist.append(newrange)
return rangelist
Example run:
getranges([2, 3, 4, 5, 12, 13, 14, 15, 16, 17])
returns:
[[2, 5], [12, 17]]
Using numpy + comprehension lists:
With numpy diff function, consequent input vector entries that their difference is not equal to one can be identified. The start and end of the input vector need to be considered.
import numpy as np
data = np.array([2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20])
d = [i for i, df in enumerate(np.diff(data)) if df!= 1]
d = np.hstack([-1, d, len(data)-1]) # add first and last elements
d = np.vstack([d[:-1]+1, d[1:]]).T
print(data[d])
Output:
[[ 2 5]
[12 17]
[20 20]]
Note: The request that individual numbers should be treated differently, (returned as individual, not ranges) was omitted. This can be reached by further post-processing the results. Usually this will make things more complex without gaining any benefit.
One-liner in Python 2.7 if interested:
x = [2, 3, 6, 7, 8, 14, 15, 19, 20, 21]
d = iter(x[:1] + sum(([i1, i2] for i1, i2 in zip(x, x[1:] + x[:1]) if i2 != i1+1), []))
print zip(d, d)
>>> [(2, 3), (6, 8), (14, 15), (19, 21)]
A short solution that works without additional imports. It accepts any iterable, sorts unsorted inputs, and removes duplicate items:
def ranges(nums):
nums = sorted(set(nums))
gaps = [[s, e] for s, e in zip(nums, nums[1:]) if s+1 < e]
edges = iter(nums[:1] + sum(gaps, []) + nums[-1:])
return list(zip(edges, edges))
Example:
>>> ranges([2, 3, 4, 7, 8, 9, 15])
[(2, 4), (7, 9), (15, 15)]
>>> ranges([-1, 0, 1, 2, 3, 12, 13, 15, 100])
[(-1, 3), (12, 13), (15, 15), (100, 100)]
>>> ranges(range(100))
[(0, 99)]
>>> ranges([0])
[(0, 0)]
>>> ranges([])
[]
This is the same as #dansalmo's solution which I found amazing, albeit a bit hard to read and apply (as it's not given as a function).
Note that it could easily be modified to spit out "traditional" open ranges [start, end), by e.g. altering the return statement:
return [(s, e+1) for s, e in zip(edges, edges)]
I copied this answer over from another question that was marked as a duplicate of this one with the intention to make it easier findable (after I just now searched again for this topic, finding only the question here at first and not being satisfied with the answers given).
The versions by Mark Byers, Andrea Ambu, SilentGhost, Nadia Alramli, and truppo are simple and fast. The 'truppo' version encouraged me to write a version that retains the same nimble behavior while handling step sizes other than 1 (and lists as singletons elements that don't extend more than 1 step with a given step size). It is given here.
>>> list(ranges([1,2,3,4,3,2,1,3,5,7,11,1,2,3]))
[(1, 4, 1), (3, 1, -1), (3, 7, 2), 11, (1, 3, 1)]
Not the best approach , but here is my 2 cents
def getConsecutiveValues2(arr):
x = ""
final = []
end = 0
start = 0
for i in range(1,len(arr)) :
if arr[i] - arr[i-1] == 1 :
end = i
else :
print(start,end)
final.append(arr[start:end+1])
start = i
if i == len(arr) - 1 :
final.append(arr[start:end+1])
return final
x = [1,2,3,5,6,8,9,10,11,12]
print(getConsecutiveValues2(x))
>> [[1, 2, 3], [5, 6], [8, 9, 10, 11]]
This implementation works for regular or irregular steps
I needed to achieve the same thing but with the slight difference where steps can be irregular. this is my implementation
def ranges(l):
if not len(l):
return range(0,0)
elif len(l)==1:
return range(l[0],l[0]+1)
# get steps
sl = sorted(l)
steps = [i-j for i,j in zip(sl[1:],sl[:-1])]
# get unique steps indexes range
groups = [[0,0,steps[0]],]
for i,s in enumerate(steps):
if s==groups[-1][-1]:
groups[-1][1] = i+1
else:
groups.append( [i+1,i+1,s] )
g2 = groups[-2]
if g2[0]==g2[1]:
if sl[i+1]-sl[i]==s:
_=groups.pop(-2)
groups[-1][0] = i
# create list of ranges
return [range(sl[i],sl[j]+s,s) if s!=0 else [sl[i]]*(j+1-i) for i,j,s in groups]
Here's an example
from timeit import timeit
# for regular ranges
l = list(range(1000000))
ranges(l)
>>> [range(0, 1000000)]
l = list(range(10)) + list(range(20,25)) + [1,2,3]
ranges(l)
>>> [range(0, 2), range(1, 3), range(2, 4), range(3, 10), range(20, 25)]
sorted(l);[list(i) for i in ranges(l)]
>>> [0, 1, 1, 2, 2, 3, 3, 4, 5, 6, 7, 8, 9, 20, 21, 22, 23, 24]
>>> [[0, 1], [1, 2], [2, 3], [3, 4, 5, 6, 7, 8, 9], [20, 21, 22, 23, 24]]
# for irregular steps list
l = [1, 3, 5, 7, 10, 11, 12, 100, 200, 300, 400, 60, 99, 4000,4001]
ranges(l)
>>> [range(1, 9, 2), range(10, 13), range(60, 138, 39), range(100, 500, 100), range(4000, 4002)]
## Speed test
timeit("ranges(l)","from __main__ import ranges,l", number=1000)/1000
>>> 9.303160999934334e-06
Yet another solution if you expect your input to be a set:
def group_years(years):
consecutive_years = []
for year in years:
close = {y for y in years if abs(y - year) == 1}
for group in consecutive_years:
if len(close.intersection(group)):
group |= close
break
else:
consecutive_years.append({year, *close})
return consecutive_years
Example:
group_years({2016, 2017, 2019, 2020, 2022})
Out[54]: [{2016, 2017}, {2019, 2020}, {2022}]
I have an array containing an even number of integers. The array represents a pairing of an identifier and a count. The tuples have already been sorted by the identifier. I would like to merge a few of these arrays together. I have thought of a few ways to do it but they are fairly complicated and I feel there might be an easy way to do this with python.
IE:
[<id>, <count>, <id>, <count>]
Input:
[14, 1, 16, 4, 153, 21]
[14, 2, 16, 3, 18, 9]
Output:
[14, 3, 16, 7, 18, 9, 153, 21]
It would be better to store these as dictionaries than as lists (not just for this purpose, but for other use cases, such as extracting the value of a single ID):
x1 = [14, 1, 16, 4, 153, 21]
x2 = [14, 2, 16, 3, 18, 9]
# turn into dictionaries (could write a function to convert)
d1 = dict([(x1[i], x1[i + 1]) for i in range(0, len(x1), 2)])
d2 = dict([(x2[i], x2[i + 1]) for i in range(0, len(x2), 2)])
print d1
# {16: 4, 153: 21, 14: 1}
After that, you could use any of the solutions in this question to add them together. For example (taken from the first answer):
import collections
def d_sum(a, b):
d = collections.defaultdict(int, a)
for k, v in b.items():
d[k] += v
return dict(d)
print d_sum(d1, d2)
# {16: 7, 153: 21, 18: 9, 14: 3}
collections.Counter() is what you need here:
In [21]: lis1=[14, 1, 16, 4, 153, 21]
In [22]: lis2=[14, 2, 16, 3, 18, 9]
In [23]: from collections import Counter
In [24]: dic1=Counter(dict(zip(lis1[0::2],lis1[1::2])))
In [25]: dic2=Counter(dict(zip(lis2[0::2],lis2[1::2])))
In [26]: dic1+dic2
Out[26]: Counter({153: 21, 18: 9, 16: 7, 14: 3})
or :
In [51]: it1=iter(lis1)
In [52]: it2=iter(lis2)
In [53]: dic1=Counter(dict((next(it1),next(it1)) for _ in xrange(len(lis1)/2)))
In [54]: dic2=Counter(dict((next(it2),next(it2)) for _ in xrange(len(lis2)/2)))
In [55]: dic1+dic2
Out[55]: Counter({153: 21, 18: 9, 16: 7, 14: 3})
Use collections.Counter:
import itertools
import collections
def grouper(n, iterable, fillvalue=None):
args = [iter(iterable)] * n
return itertools.izip_longest(fillvalue=fillvalue, *args)
count1 = collections.Counter(dict(grouper(2, lst1)))
count2 = collections.Counter(dict(grouper(2, lst2)))
result = count1 + count2
I've used the itertools library grouper recipe here to convert your data to dictionaries, but as other answers have shown you there are more ways to skin that particular cat.
result is a Counter with each id pointing to a total count:
Counter({153: 21, 18: 9, 16: 7, 14: 3})
Counters are multi-sets and will keep track of the count of each key with ease. It feels like a much better data structure for your data. They support summing, as used above, for example.
All of the previous answers look good, but I think that the JSON blob should be properly formed to begin with or else (from my experience) it can cause some serious problems down the road during debugging etc. In this case with id and count as the fields, the JSON should look like
[{"id":1, "count":10}, {"id":2, "count":10}, {"id":1, "count":5}, ...]
Properly formed JSON like that is much easier to deal with, and probably similar to what you have coming in anyway.
This class is a bit general, but certainly extensible
from itertools import groupby
class ListOfDicts():
def init_(self, listofD=None):
self.list = []
if listofD is not None:
self.list = listofD
def key_total(self, group_by_key, aggregate_key):
""" Aggregate a list of dicts by a specific key, and aggregation key"""
out_dict = {}
for k, g in groupby(self.list, key=lambda r: r[group_by_key]):
print k
total=0
for record in g:
print " ", record
total += record[aggregate_key]
out_dict[k] = total
return out_dict
if __name__ == "__main__":
z = ListOfDicts([ {'id':1, 'count':2, 'junk':2},
{'id':1, 'count':4, 'junk':2},
{'id':1, 'count':6, 'junk':2},
{'id':2, 'count':2, 'junk':2},
{'id':2, 'count':3, 'junk':2},
{'id':2, 'count':3, 'junk':2},
{'id':3, 'count':10, 'junk':2},
])
totals = z.key_total("id", "count")
print totals
Which gives
1
{'count': 2, 'junk': 2, 'id': 1}
{'count': 4, 'junk': 2, 'id': 1}
{'count': 6, 'junk': 2, 'id': 1}
2
{'count': 2, 'junk': 2, 'id': 2}
{'count': 3, 'junk': 2, 'id': 2}
{'count': 3, 'junk': 2, 'id': 2}
3
{'count': 10, 'junk': 2, 'id': 3}
{1: 12, 2: 8, 3: 10}