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Split List By Value and Keep Separators
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Is there an easy way to split the list l below into 3 list. I want to cut the list when the sequence starts over. So every list should start with 1.
l= [1, 2, 3, 4, 5, 1, 2, 3, 4, 1, 2, 3, 4]
l1 = [1, 2, 3,4, 5]
l2=[1,2,3,4]
l3=[1,2,3,4]
My original thought was to look at the lead value and implement a condition inside a for loop that would cut the list when x.lead < x. But how do I use lead and lag when using lists in python?
NumPy solution
import numpy as np
l = [1, 2, 3, 4, 5, 1, 2, 3, 4, 1, 2, 3, 4]
parts = [list(i) for i in np.split(l,np.flatnonzero(np.diff(l)-1)+1)]
print(parts)
output
[[1, 2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4]]
Explanation: I first find differences between adjacent elements using numpy.diff, then subtract 1 to be able to use numpy.flatnonzero to find where difference is other than 1, add 1 (note that numpy.diff output length is input length minus 1) to get indices for use in numpy.split, eventually convert it to list, as otherwise you would end with numpy.arrays
What about this:
l = [1, 2, 3, 4, 5, 1, 2, 3, 4, 1, 2, 3, 4]
one_indices = [i for i, e in enumerate(l) if e == 1]
slices = []
for count, item in enumerate(one_indices):
if count == len(one_indices) - 1:
slices.append((item, None))
else:
slices.append((item, one_indices[count + 1]))
sequences = [l[x[0] : x[1]] for x in slices]
print(sequences)
Out:
[[1, 2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4]]
Another way without numpy,
l= [1, 2, 3, 4, 5, 1, 2, 3, 4, 1, 2, 3, 4]
start = 0
newlist = []
for i,v in enumerate(l):
if i!=0 and v==1:
newlist.append(l[start:i])
start = i
newlist.append(l[start:i+1])
print(newlist)
Working Demo: https://rextester.com/RYCV85570
I want to weave two lists and output all the possible results.
For example,
input: two lists l1 = [1, 2], l2 = [3, 4]
output: [1, 2, 3, 4], [1, 3, 2, 4], [1, 3, 4, 2], [3, 1, 2, 4], [3, 1, 4, 2], [3, 4, 1, 2]
Note: I need to keep the order in each list (e.g. 1 is always before 2, and 3 is always before 4)
The way I am solving this is by removing the head from one list, recursing, and then doing the same thing with the other list. The code is below:
all_possibles = []
def weaveLists(first, second, added):
if len(first) == 0 or len(second) == 0:
res = added[:]
res += first[:]
res += second[:]
all_possibles.append(res)
return
cur1 = first[0]
added.append(cur1)
first = first[1:]
weaveLists(first, second, added)
added = added[:-1]
first = [cur1] + first
cur2 = second[0]
added.append(cur2)
second = second[1:]
weaveLists(first, second, added)
added = added[:-1]
second = [cur2] + second
weaveLists([1, 2], [3, 4], [])
print(all_possibles)
The result I got is:
[[1, 2, 3, 4], [1, 3, 2, 4], [1, 3, 4, 2], [1, 3, 1, 2, 4], [1, 3, 1, 4, 2], [1, 3, 1, 4, 1, 2]]
I couldn't figure out why for the last three lists, the heading 1 from the first list is not removed.
Can anyone help? Thanks!
The reason you get those unexpected results is that you mutate added at this place:
added.append(cur1)
...this will affect the caller's added list (unintentionally). While the "undo" operation is not mutating the list:
added = added[:-1]
This creates a new list, and therefore this "undo" action does not roll back the change in the list of the caller.
The easy fix is to replace the call to append with:
added = added + [cur1]
And the same should happen in the second block.
It is easier if you pass the new values for the recursive call on-the-fly, and replace those two code blocks with just:
weaveLists(first[1:], second, added + [first[0]])
weaveLists(first, second[1:], added + [second[0]])
Here is another way to do it: we generate the possible indices of the items of the first list inside the weaved list, and fill the list accordingly.
We can generate the indices with itertools.combinations: it's the combinations of the indices of the weaved list, taking len(first_list) of them each time.
from itertools import combinations
def weave(l1, l2):
total_length = len(l1) + len(l2)
# indices at which to put items from l1 in the weaved output
for indices in combinations(range(total_length), r=len(l1)):
out = []
it1 = iter(l1)
it2 = iter(l2)
for i in range(total_length):
if i in indices:
out.append(next(it1))
else:
out.append(next(it2))
yield out
Sample run:
l1 = [1, 2]
l2 = [3, 4]
for w in weave(l1, l2):
print(w)
[1, 2, 3, 4]
[1, 3, 2, 4]
[1, 3, 4, 2]
[3, 1, 2, 4]
[3, 1, 4, 2]
[3, 4, 1, 2]
Another sample run with a longer list:
l1 = [1, 2]
l2 = [3, 4, 5]
for w in weave(l1, l2):
print(w)
[1, 2, 3, 4, 5]
[1, 3, 2, 4, 5]
[1, 3, 4, 2, 5]
[1, 3, 4, 5, 2]
[3, 1, 2, 4, 5]
[3, 1, 4, 2, 5]
[3, 1, 4, 5, 2]
[3, 4, 1, 2, 5]
[3, 4, 1, 5, 2]
[3, 4, 5, 1, 2]
I want to create a date list that- contains the list of dates of one week but each date should be there 4 times in that list. like this:
[1-jan-2018, 1-jan-2018, 1-jan-2018, 1-jan-2018, 2-jan-2018, 2-jan-2018, 2-jan-2018, 2-jan-2018, 3-jan-2018, 3-jan-2018, 3-jan-2018, 3-jan-2018, 4-jan-2018, 4-jan-2018, 4-jan-2018, 4-jan-2018, 5-jan-2018, 5-jan-2018, 5-jan-2018, 5-jan-2018, 6-jan-2018, 6-jan-2018, 6-jan-2018, 6-jan-2018,7-jan-2018, 7-jan-2018, 7-jan-2018, 7-jan-2018]
I don't exactly have the idea how to do it but here is my attempt:
import pandas as pd
timeSeries = list(pd.date_range(start='1/1/2020', end='7/1/2020'))
print(timeSeries)
This will just create the list of dates of one week but I want the answer in the above format. Can someone please help?
How to duplicate items in a list
A solution is create various list with each element of your primary list repeated N time. In this example, I will duplicated each element four times, so:
old_list = [1,2,3,4]
# [i,i,i,i] will clone each item four times.
new_list = list([i,i,i,i] for i in old_list)
# new_list = [[1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], [4, 4, 4, 4]]
But, now you will have a list of lists, so you need to transform that result into a list of elements, this operation is called flat. In order to do this in python, you can use the itertools.chain.
import itertools
old_list = [1,2,3,4]
# [i,i,i,i] will clone each item four times.
new_list = list([i,i,i,i] for i in old_list)
# new_list = [[1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], [4, 4, 4, 4]]
new_list_flatten = list(itertools.chain(*new_list))
# new_list_flatten = [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4]
You can avoid the usage of * operation by calling itertools.chain.from_iterable:
import itertools
old_list = [1,2,3,4]
# [i,i,i,i] will clone each item four times.
new_list = list([i,i,i,i] for i in old_list)
# new_list = [[1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], [4, 4, 4, 4]]
new_list_flatten = list(itertools.chain.from_iterable(new_list))
# new_list_flatten = [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4]
Therefore, this is a code that do what you want:
import itertools
import pandas as pd
time_series = list(pd.date_range(start='1/1/2020', end='7/1/2020'))
newseries = list( itertools.chain.from_iterable((i,i,i,i) for i in time_series) )
print(newseries)
So there is a list which contains
'3,2,5,4,1','3,1,2,5,4','2,5,1,4,3'
These numbers are part of the same list, HOWEVER they are strings in a list(ie. list 1)
and from this, you say that for the "first row", 3 occurs at position 0, 2 occurs at position 1, 5 at 2 etc.
For the "second row", 3 occurs at position 0, 1 occurs at position 1, 2 occurs at position 2 etc.
I would like to create a loop or anything at all (besides using imported functions) to create a final list which looks like
0: [3, 3, 2]
1: [2, 1, 5]
2: [5, 2, 1]
3: [4, 5, 4]
4: [1, 4, 3]
Transposition of a two-dimensional list can simply be done using zip()
In [1]: l = [[3,2,5,4,1],
...: [3,1,2,5,4],
...: [2,5,1,4,3]]
In [2]: t = list(zip(*l))
In [3]: t
Out[3]: [(3, 3, 2), (2, 1, 5), (5, 2, 1), (4, 5, 4), (1, 4, 3)]
To output that in the format described above:
In [4]: for n,line in enumerate(t):
...: print("{}: {}".format(n, list(line)))
...:
0: [3, 3, 2]
1: [2, 1, 5]
2: [5, 2, 1]
3: [4, 5, 4]
4: [1, 4, 3]
Single line of code using Dictionary comprehension and List comprehension :
>>> { col:[row[col] for row in l] for col in range(len(l[0])) }
=> {0: [3, 3, 2], 1: [2, 1, 5], 2: [5, 2, 1], 3: [4, 5, 4], 4: [1, 4, 3]}
#driver values :
IN : l = [[3,2,5,4,1],
[3,1,2,5,4],
[2,5,1,4,3]]
NOTE to OP : what your output suggests is a Dictionary by looking at its structure. A list cannot be defined in the same manner.
EDIT : Since the OP's list is a list of strings, first convert that to a list of int using map and then continue as above.
>>> l = ['3,2,5,4,1', '3,1,2,5,4', '2,5,1,4,3']
>>> l = [list(map(int,s.split(','))) for s in l]
>>> l
=> [[3, 2, 5, 4, 1], [3, 1, 2, 5, 4], [2, 5, 1, 4, 3]]
I have a list [[1, 2, 7], [1, 2, 3], [1, 2, 3, 7], [1, 2, 3, 5, 6, 7]] and I need [1,2,3,7] as final result (this is kind of reverse engineering). One logic is to check intersections -
while(i<dlistlen):
j=i+1
while(j<dlistlen):
il = dlist1[i]
jl = dlist1[j]
tmp = list(set(il) & set(jl))
print tmp
#print i,j
j=j+1
i=i+1
this is giving me output :
[1, 2]
[1, 2, 7]
[1, 2, 7]
[1, 2, 3]
[1, 2, 3]
[1, 2, 3, 7]
[]
Looks like I am close to getting [1,2,3,7] as my final answer, but can't figure out how. Please note, in the very first list (([[1, 2, 7], [1, 2, 3], [1, 2, 3, 7], [1, 2, 3, 5, 6, 7]] )) there may be more items leading to one more final answer besides [1,2,3,4]. But as of now, I need to extract only [1,2,3,7] .
Please note, this is not kind of homework, I am creating own clustering algorithm that fits my need.
You can use the Counter class to keep track of how often elements appear.
>>> from itertools import chain
>>> from collections import Counter
>>> l = [[1, 2, 7], [1, 2, 3], [1, 2, 3, 7], [1, 2, 3, 5, 6, 7]]
>>> #use chain(*l) to flatten the lists into a single list
>>> c = Counter(chain(*l))
>>> print c
Counter({1: 4, 2: 4, 3: 3, 7: 3, 5: 1, 6: 1})
>>> #sort keys in order of descending frequency
>>> sortedValues = sorted(c.keys(), key=lambda x: c[x], reverse=True)
>>> #show the four most common values
>>> print sortedValues[:4]
[1, 2, 3, 7]
>>> #alternatively, show the values that appear in more than 50% of all lists
>>> print [value for value, freq in c.iteritems() if float(freq) / len(l) > 0.50]
[1, 2, 3, 7]
It looks like you're trying to find the largest intersection of two list elements. This will do that:
from itertools import combinations
# convert all list elements to sets for speed
dlist = [set(x) for x in dlist]
intersections = (x & y for x, y in combinations(dlist, 2))
longest_intersection = max(intersections, key=len)