Comparing lists with their indices and content in Python - python

I have a list of numbers as
N = [13, 14, 15, 25, 27, 31, 35, 36, 43]
After some calculations, for each element in N, I get the following list as the answers.
ndlist = [4, 30, 0, 42, 48, 4, 3, 42, 3]
That is, for the first index in N (which is 13), my answer is 4 in ndlist.
For some indices in N, I get the same answer in ndlist. For example, when N= 13 and 31, the answer is 4 in ndlist.
I need to find the numbers in N (13 and 31 in my example) such that they have the same answer in ndlist.
Can someone help me to that?

You can use a defaultdict and put those into a list keyed by the answer like:
Code:
N = [13, 14, 15, 25, 27, 31, 35, 36, 43]
ndlist = [4, 30, 0, 42, 48, 4, 3, 42, 3]
from collections import defaultdict
answers = defaultdict(list)
for n, answer in zip(N, ndlist):
answers[answer].append(n)
print(answers)
print([v for v in answers.values() if len(v) > 1])
Results:
defaultdict(<class 'list'>, {4: [13, 31], 30: [14],
0: [15], 42: [25, 36], 48: [27], 3: [35, 43]})
[[13, 31], [25, 36], [35, 43]]

Here is a way using only a nested list comprehension:
[N[idx] for idx, nd in enumerate(ndlist) if nd in [i for i in ndlist if ndlist.count(i)>1]]
#[13, 25, 31, 35, 36, 43]
To explain: the inner list comprehension ([i for i in ndlist if ndlist.count(i)>1]) gets all duplicate values in ndlist, and the rest of the list comprehension extracts the corresponding values in N where those values are found in ndlist

Related

Add new element in the next sublist depending in if it has been added or not (involves also a dictionary problem) python

Community of Stackoverflow:
I'm trying to create a list of sublists with a loop based on a random sampling of values of another list; and each sublist has the restriction of not having a duplicate or a value that has already been added to a prior sublist.
Let's say (example) I have a main list:
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
#I get:
[[1,13],[4,1],[8,13]]
#I WANT:
[[1,13],[4,9],[8,14]] #(no duplicates when checking previous sublists)
The real code that I thought it would work is the following (as a draft):
matrixvals=list(matrix.index.values) #list where values are obtained
lists=[[]for e in range(0,3)] #list of sublists that I want to feed
vls=[] #stores the values that have been added to prevent adding them again
for e in lists: #initiate main loop
for i in range(0,5): #each sublist will contain 5 different random samples
x=random.sample(matrixvals,1) #it doesn't matter if the samples are 1 or 2
if any(x) not in vls: #if the sample isn't in the evaluation list
vls.extend(x)
e.append(x)
else: #if it IS, then do a sample but without those already added values (line below)
x=random.sample([matrixvals[:].remove(x) for x in vls],1)
vls.extend(x)
e.append(x)
print(lists)
print(vls)
It didn't work as I get the following:
[[[25], [16], [15], [31], [17]], [[4], [2], [13], [42], [13]], [[11], [7], [13], [17], [25]]]
[25, 16, 15, 31, 17, 4, 2, 13, 42, 13, 11, 7, 13, 17, 25]
As you can see, number 13 is repeated 3 times, and I don't understand why
I would want:
[[[25], [16], [15], [31], [17]], [[4], [2], [13], [42], [70]], [[11], [7], [100], [18], [27]]]
[25, 16, 15, 31, 17, 4, 2, 13, 42, 70, 11, 7, 100, 18, 27] #no dups
In addition, is there a way to convert the sample.random results as values instead of lists? (to obtain):
[[25,16,15,31,17]], [4, 2, 13, 42,70], [11, 7, 100, 18, 27]]
Also, the final result in reality isn't a list of sublists, actually is a dictionary (the code above is a draft attempt to solve the dict problem), is there a way to obtain that previous method in a dict? With my present code I got the next results:
{'1stkey': {'1stsubkey': {'list1': [41,
40,
22,
28,
26,
14,
41,
15,
40,
33],
'list2': [41, 40, 22, 28, 26, 14, 41, 15, 40, 33],
'list3': [41, 40, 22, 28, 26, 14, 41, 15, 40, 33]},
'2ndsubkey': {'list1': [21,
7,
31,
12,
8,
22,
27,...}
Instead of that result, I would want the following:
{'1stkey': {'1stsubkey': {'list1': [41,40,22],
'list2': [28, 26, 14],
'list3': [41, 15, 40, 33]},
'2ndsubkey': {'list1': [21,7,31],
'list2':[12,8,22],
'list3':[27...,...}#and so on
Is there a way to solve both list and dict problem? Any help will be very appreciated; I can made some progress even only with the list problem
Thanks to all
I realize you may be more interested in finding out why your particular approach isn't working. However, if I've understood your desired behavior, I may be able to offer an alternative solution. After posting my answer, I will take a look at your attempt.
random.sample lets you sample k number of items from a population (collection, list, whatever.) If there are no repeated elements in the collection, then you're guaranteed to have no repeats in your random sample:
from random import sample
pool = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
num_samples = 4
print(sample(pool, k=num_samples))
Possible output:
[9, 11, 8, 7]
>>>
It doesn't matter how many times you run this snippet, you will never have repeated elements in your random sample. This is because random.sample doesn't generate random objects, it just randomly picks items which already exist in a collection. This is the same approach you would take when drawing random cards from a deck of cards, or drawing lottery numbers, for example.
In your case, pool is the pool of possible unique numbers to choose your sample from. Your desired output seems to be a list of three lists, where each sublist has two samples in it. Rather than calling random.sample three times, once for each sublist, we should call it once with k=num_sublists * num_samples_per_sublist:
from random import sample
pool = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
num_sublists = 3
samples_per_sublist = 2
num_samples = num_sublists * samples_per_sublist
assert num_samples <= len(pool)
print(sample(pool, k=num_samples))
Possible output:
[14, 10, 1, 8, 6, 3]
>>>
OK, so we have six samples rather than four. No sublists yet. Now you can simply chop this list of six samples up into three sublists of two samples each:
from random import sample
pool = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
num_sublists = 3
samples_per_sublist = 2
num_samples = num_sublists * samples_per_sublist
assert num_samples <= len(pool)
def pairwise(iterable):
yield from zip(*[iter(iterable)]*samples_per_sublist)
print(list(pairwise(sample(pool, num_samples))))
Possible output:
[(4, 11), (12, 13), (8, 15)]
>>>
Or if you really want sublists, rather than tuples:
def pairwise(iterable):
yield from map(list, zip(*[iter(iterable)]*samples_per_sublist))
EDIT - just realized that you don't actually want a list of lists, but a dictionary. Something more like this? Sorry I'm obsessed with generators, and this isn't really easy to read:
keys = ["1stkey"]
subkeys = ["1stsubkey", "2ndsubkey"]
num_lists_per_subkey = 3
num_samples_per_list = 5
num_samples = num_lists_per_subkey * num_samples_per_list
min_sample = 1
max_sample = 50
pool = list(range(min_sample, max_sample + 1))
def generate_items():
def generate_sub_items():
from random import sample
samples = sample(pool, k=num_samples)
def generate_sub_sub_items():
def chunkwise(iterable, n=num_samples_per_list):
yield from map(list, zip(*[iter(iterable)]*n))
for list_num, chunk in enumerate(chunkwise(samples), start=1):
key = f"list{list_num}"
yield key, chunk
for subkey in subkeys:
yield subkey, dict(generate_sub_sub_items())
for key in keys:
yield key, dict(generate_sub_items())
print(dict(generate_items()))
Possible output:
{'1stkey': {'1stsubkey': {'list1': [43, 20, 4, 27, 2], 'list2': [49, 44, 18, 8, 37], 'list3': [19, 40, 9, 17, 6]}, '2ndsubkey': {'list1': [43, 20, 4, 27, 2], 'list2': [49, 44, 18, 8, 37], 'list3': [19, 40, 9, 17, 6]}}}
>>>

Create a dictionary from a list of key and multi list of values in Python

I have a list of key:
list_date = ["MON", "TUE", "WED", "THU","FRI"]
I have many lists of values that created by codes below:
list_value = list()
for i in list(range(5, 70, 14)):
list_value.append(list(range(i, i+10, 3)))
Rules created that:
first number is 5, a list contains 4 items has value equal x = x + 3, and so on [5, 8, 11,1 4]
the first number of the second list equal: x = 5 + 14, and value inside still as above x = x +3
[[5, 8, 11, 14], [19, 22, 25, 28], [33, 36, 39, 42], [47, 50, 53, 56], [61, 64, 67, 70]]
I expect to obtain a dict like this:
collections = {"MON":[5, 8, 11, 14], "TUE" :[19, 22, 25, 28], "WED":[33, 36, 39, 42], "THU":[47, 50, 53, 56], "FRI":[61, 64, 67, 70]}
Then, I used:
zip_iterator = zip(list_date, list_value)
collections = dict(zip_iterator)
To get my expected result.
I tried another way like using lambda function
for i in list(range(5, 70, 14)):
list_value.append(list(range(i,i+10,3)))
couple_start_end[lambda x: x in list_date] = list(range(i, i + 10, 3))
And the output is:
{<function <lambda> at 0x000001BF7F0711F0>: [5, 8, 11, 14], <function <lambda> at 0x000001BF7F071310>: [19, 22, 25, 28], <function <lambda> at 0x000001BF7F071280>: [33, 36, 39, 42], <function <lambda> at 0x000001BF7F0710D0>: [47, 50, 53, 56], <function <lambda> at 0x000001BF7F0890D0>: [61, 64, 67, 70]}
I want to ask there is any better solution to create lists of values with the rules above? and create the dictionary collections without using the zip method?
Thank you so much for your attention and participation.
Sure, you can use enumerate but I wouldn't say it is in anyway better or worse than the zip based solution:
collections = {}
for idx, key in enumerate(list_keys):
collections[key] = list_value[idx]
print(collections)
Output:
{'MON': [5, 8, 11, 14], 'TUE': [19, 22, 25, 28], 'WED': [33, 36, 39, 42], 'THU': [47, 50, 53, 56], 'FRI': [61, 64, 67, 70]}
Further, you don't need to create the value list separately, you can create the dictionary as you go along:
list_keys = ["MON", "TUE", "WED", "THU","FRI"]
collections = {}
for idx, start in enumerate(range(5, 70, 14)):
collections[list_keys[idx]] = [i for i in range(start, start+10, 3)]
print(collections)

Print list in specified range Python

I'm new to Python and I have this problem
I have a list of numbers like this:
n = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
I want to print from 11 to 37, that means the output = 11, 13,.... 37.
I tried to print(n[11:37]) but of course it will print [37, 41, 43, 47]
because that is range index.
Any ideas or does Python have any built-in method for this ?
This should do the job...
n = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
n.sort()
mylist = [x for x in n if x in range(11, 38)]
print(mylist)
Want to print that as comma separated string:
print(mylist.strip('[]'))
This will work. (Assuming list is sorted)
print n[n.index(11): n.index(37)+1]
Output:
[11, 13, 17, 19, 23, 29, 31, 37]
Considering your list is ordered and it has no duplicates:
n = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
print(",".join(map(str,n[n.index(11): n.index(37)+1])))
Here you have a live example
Using numpy:
import numpy as np
narr = np.array(n)
m = (narr >= 11) & (narr <= 37)
for v in narr[m]:
print(v)
# or, to get rid of the loop:
print('\n'.join(map(str, narr[m])))
it pretty simple, since your list is already sorted you can write
my_list = [x for x in n if x in range(11, 38)]
print(*my_list)
what the '*' does is that it unpacks the array into individual elements, a term known as unpacking.This will produce the actual result you wanted and not an array
If your data is sorted, you can use a generator expression with either a range object or chained comparisons:
n = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
print(*(i for i in n if i in range(11, 38)), sep=', ')
print(*(i for i in n if 11 <= i <= 37), sep=', ')
If your data is unsorted and you can use indices of the first occurrences of each value, you can slice your list:
print(*n[n.index(11): n.index(37)+1], sep=', ')
Result with the data you have provided:
11, 13, 17, 19, 23, 29, 31, 37

Periodically slice an list/array

Suppose i have a a = range(1,51). How can i slice a to create a new list that look like this:
[1,2,3,11,12,13,21,22,23,31,32,33,41,42,43]
Is there a pythonic way that can help me do this without writing function?
I know that [start:stop:step] for periodically slicing one element but i'm not sure if i'm missing something obvious.
EDIT: The suggested duplicate question/answer is not the same as mine question. I simply asked to slice/extract periodically elements from a larger list/array. The suggested duplicate modifies elements of existing array.
Another option you can go with logical vector subsetting, something like:
a[(a - 1) % 10 < 3]
# array([ 1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43])
(a - 1) % 10 finds the remainder of array by 10 (period); and (a - 1) % 10 < 3 gives a logical vector which gives true for the first three elements of every ten elements.
What you want is more complicated than a simple slice, so you're going to need some kind of (likely fairly simple) function to do it. I'd look at using zip to combine multiple slices, something like:
reduce(lambda a,b:a+b, map(list, zip(a[1::10], a[2::10], a[3::10])))
Given:
>>> li=range(1,52)
You can do:
>>> [l for sl in [li[i:i+3] for i in range(0,len(li),10)] for l in sl]
[1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43, 51]
Or, if you want only full sublists:
>>> [l for sl in [li[i:i+3] for i in range(0,len(li),10)] for l in sl if len(sl)==3]
[1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43]
Or, given:
>>> li=range(1,51)
Then you do not need to test sublists:
>>> [l for sl in [li[i:i+3] for i in range(0,len(li),10)] for l in sl]
[1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43]
Psidom's answer's index math can be adapted to a list comprehension too
a = range(1,51)
[n for n in a if (n - 1) % 10 < 3]
Out[23]: [1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43]

Python: Renumerate elements in multiple lists

Suppose I have a dictionary with lists as follows:
{0: [31, 32, 58, 59], 1: [31, 32, 12, 13, 37, 38], 2: [12, 13]}
I am trying to obtain the following one from it:
{0: [1, 2, 3, 4], 1: [1, 2, 5, 6, 7, 8], 2: [5, 6]}
So I renumerate all the entries in order of occurence but skipping those that were already renumerated.
What I have now is a bunch of for loops going back and forth which works, but doesn't look good at all, could anyone please tell me the way it should be done in Python 2.7?
Thank you
import operator
data = {0: [31, 32, 58, 59], 1: [31, 32, 12, 13, 37, 38], 2: [12, 13]}
# the accumulator is the new dict with renumbered values combined with a list of renumbered numbers so far
# item is a (key, value) element out of the original dict
def reductor(acc, item):
(out, renumbered) = acc
(key, values) = item
def remapper(v):
try:
x = renumbered.index(v)
except ValueError:
x = len(renumbered)
renumbered.append(v)
return x
# transform current values to renumbered values
out[key] = map(remapper, values)
# return output and updated list of renumbered values
return (out, renumbered)
# now reduce the original data
print reduce(reductor, sorted(data.iteritems(), key=operator.itemgetter(0)), ({}, []))
If you're not worried about memory or speed you can use an intermediate dictionary to map the new values:
a = {0: [31, 32, 58, 59], 1: [31, 32, 12, 13, 37, 38], 2: [12, 13]}
b = {}
c = {}
for key in sorted(a.keys()):
c[key] = [b.setdefault(val, len(b)+1) for val in a[key]]
Just use a function like this:
def renumerate(data):
ids = {}
def getid(val):
if val not in ids:
ids[val] = len(ids) + 1
return ids[val]
return {k : map(getid, data[k]) for k in sorted(data.keys())}
Example
>>> data = {0: [31, 32, 58, 59], 1: [31, 32, 12, 13, 37, 38], 2: [12, 13]}
>>> print renumerate(data)
{0: [1, 2, 3, 4], 1: [1, 2, 5, 6, 7, 8], 2: [5, 6]}
data = {0: [31, 32, 58, 59], 1: [31, 32, 12, 13, 37, 38], 2: [12, 13]}
from collections import defaultdict
numbered = defaultdict(lambda: len(numbered)+1)
result = {key: [numbered[v] for v in val] for key, val in sorted(data.iteritems(), key=lambda item: item[0])}
print result

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