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Index array based on value limits of another

Let's say I have an array (or even a list) that looks like:
tmp_data = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
And then I have another ray that are distance values:
dist_data = [ 15.625 46.875 78.125 109.375 140.625 171.875 203.125 234.375 265.625 296.875]
Now, say I want to create a threshold of distance that I would like to perform an operation on from tmp_data. For this example, let's just take the max value. And let's set the threshold distance to 100. What I would like to do is take the n number of elements every 100 distance units and replace all elements in that with the maximum value in that small array. For example: I would want the final output to be
max_tmp_data_100 = [2,2,2,5,5,5,8,8,8,9]
This is because the first 3 elements in dist_data are below 100, so we take the first three elements of tmp_data (0,1,2), and get the maximum of this and replace all elements in there with that value, 2
Then, the next set of data that would be below the next 100 value would be
tmp_dist_array_100 = [109.375 140.625 171.875]
tmp_data_100 = [3,4,5]
max_tmp_data_100 = [5,5,5]
(append to [2,2,2])
I have come up with the following:
# Initialize
final_array = []
d_array = []
idx = 1
for i in range(0,10):
if dist_data[i] < idx * final_res:
d_array.append(tmp_data[i])
elif dist_data[i] > idx * final_res:
# Now get the values
max_val = np.amax(d_array)
new_array = np.ones(len(d_array)) * max_val
final_array.extend(new_array)
idx = idx + 1
But the outcome is
[2.0, 2.0, 2.0, 5.0, 5.0, 5.0, 5.0, 5.0]
When it should be [2,2,2,5,5,5,8,8,8,9]

With numpy:
import numpy as np
cdist_data = [15.625, 46.875, 78.125, 109.375, 140.625, 171.875, 203.125, 234.375,265.625, 296.875]
cut = 100
a = np.array(dist_data)
vals = np.searchsorted(a, np.r_[cut:a.max() + cut:cut]) - 1
print(vals[(a/cut).astype(int)])
It gives:
[2 2 2 5 5 5 9 9 9 9]

You can do with groupby
from itertools import groupby
dist_data = [ 15.625, 46.875 ,78.125 ,109.375 ,140.625 ,171.875 ,203.125 ,234.375, 265.625 ,296.875]
tmp_data = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
result = []
index_list = [[dist_data.index(i) for i in l]
for k, l in groupby(dist_data, key=lambda x:x//100)]
for i in tmp_data:
for lst in index_list:
if i in lst:
result.append(max(lst))
print(result)
# [2, 2, 2, 5, 5, 5, 9, 9, 9, 9]
A per your requirements last 4 elements will comes under next threshold value, the max of last 4 element is 9.

How to get ranks from a sample in a list of values?

I'm new with Python and have a quite simple problem on paper but difficult to me in Python.
I have two samples of values (which are lists) :
X = [2, 2, 4, 6]
Y = [1, 3, 4, 5]
I have a concatenated list which is sorted as
Z = [ 1 , 2 , 2 , 3 , 4 , 4 , 5 , 6]
#rank: 1 2.5 4 5.5 7 8
I would like to get the sum of ranks of X values in Z. For this example, the ranks of 2, 2, 4 and 6 in Z are 2.5 + 2.5 + 5.5 + 8 = 18.5
(ranks of Y values in Z are 1 + 4 + 5.5 + 7 = 17.5)
Here is what I've done but it doesn't work with these lists X and Y (it works if each value appears only one time)
def funct(X, Z):
rank = []
for i in range(len(Z)):
for j in range(len(X)):
if Z[i] == X[j]:
rank = rank + [(i+1)]
print(sum(rank))
return
I would like to solve my problem with not too much complicated functions (only loops and quite easy ways to get a solution).

You can use a dictionary to keep track of the rank sums and counts once you've sorted the combined list.
X = [2, 2, 4, 6]
Y = [1, 3, 4, 5]
Z = sorted(X + Y)
ranksum = {}
counts = {}
for i, v in enumerate(Z):
ranksum[v] = ranksum.get(v, 0) + (i + 1) # Add
counts[v] = counts.get(v, 0) + 1 # Increment count
Then, when you want to look up the rank of an element, you need ranksum[v] / count[v].
r = [ranksum[x] / counts[x] for x in X]
print(r)
# Out: [2.5, 2.5, 5.5, 8]

Here's a solution for how to build the list of ranks:
X = ...
Y = ...
Z = sorted(X + Y)
rank = [1]
z = Z[:1]
for i, e in enumerate(Z[1:], start=2):
if e == z[-1]:
rank[-1] += 0.5
else:
rank.append(i)
z.append(e)
Now you can convert that into a dictionary:
ranks = dict(zip(z, rank))
That will make lookup easier:
sum(ranks[e] for e in X)

Here's another option where you build a dictionary of the rank indexes and then create a rank dictionary from there:
from collections import defaultdict
X = [2, 2, 4, 6]
Y = [1, 3, 4, 5]
Z = sorted(X + Y)
rank_indexes = defaultdict(lambda: [])
for i,v in enumerate(Z):
rank_indexes[v].append(i+1)
ranks = {k:(sum(v)/len(v)) for (k,v) in rank_indexes.items()}
print("Sum of X ranks:", sum([ranks[v] for v in X]))
print("Sum of Y ranks:", sum([ranks[v] for v in Y]))
Output:
Sum of X ranks: 18.5
Sum of Y ranks: 17.5
You can do the same thing without defaultdict, but it's slightly slower and I'd argue less Pythonic:
rank_indexes = {}
for i,v in enumerate(Z):
rank_indexes.setdefault(v, []).append(i+1)
ranks = {k:(sum(v)/len(v)) for (k,v) in rank_indexes.items()}

python: Create new lists by summing values of multiple lists

I want to create a new list that is the sum of the columns of the previous lists.
I have a lot of different lists and I would like to sum up all of the different lists in the most efficient way possible. Below is an example of the issue I am trying to solve:
list[0] = [2,4,1,6,7]
list[1] = [3,1,2,11,0]
list[2] = [2,4,2,2,1]
...
list[999] = [4,2,5,6,7]
The newlist would then look something like this:
Newlist = [1340,1525,675,1825,895]
What would be the best way to create the new list.

What do you think of this solution:
import numpy as np
a = list()
a.append([2, 4, 1, 6, 7]) # a[0]
a.append([3, 1, 2, 11, 0]) # a[1]
a.append([2, 4, 2, 2, 1]) # a[2]
# 1st solution
rslt_1 = a[0]
for _ in range(1, len(a)):
rslt_1 = np.add(rslt_1, a[_])
# 2nd solution
rslt_2 = np.sum(a, axis=0)
print("Rslt_1:", rslt_1)
print("Rslt_2:", rslt_2)
Returns:
Rslt_1: [ 7 9 5 19 8]
Rslt_2: [ 7 9 5 19 8]

Python sort array by another positions array

Assume I have two arrays, the first one containing int data, the second one containing positions
a = [11, 22, 44, 55]
b = [0, 1, 10, 11]
i.e. I want a[i] to be be moved to position b[i] for all i. If I haven't specified a position, then insert a -1
i.e
sorted_a = [11, 22,-1,-1,-1,-1,-1,-1,-1,-1, 44, 55]
^ ^ ^ ^
0 1 10 11
Another example:
a = [int1, int2, int3]
b = [5, 3, 1]
sorted_a = [-1, int3, -1, int2, -1, int1]
Here's what I've tried:
def sort_array_by_second(a, b):
sorted = []
for e1 in a:
sorted.appendAt(b[e1])
return sorted
Which I've obviously messed up.

Something like this:
res = [-1]*(max(b)+1) # create a list of required size with only -1's
for i, v in zip(b, a):
res[i] = v
The idea behind the algorithm:
Create the resulting list with a size capable of holding up to the largest index in b
Populate this list with -1
Iterate through b elements
Set elements in res[b[i]] with its proper value a[i]
This will leave the resulting list with -1 in every position other than the indexes contained in b, which will have their corresponding value of a.

I would use a custom key function as an argument to sort. This will sort the values according to the corresponding value in the other list:
to_be_sorted = ['int1', 'int2', 'int3', 'int4', 'int5']
sort_keys = [4, 5, 1, 2, 3]
sort_key_dict = dict(zip(to_be_sorted, sort_keys))
to_be_sorted.sort(key = lambda x: sort_key_dict[x])
This has the benefit of not counting on the values in sort_keys to be valid integer indexes, which is not a very stable thing to bank on.

>>> a = ["int1", "int2", "int3", "int4", "int5"]
>>> b = [4, 5, 1, 2, 3]
>>> sorted(a, key=lambda x, it=iter(sorted(b)): b.index(next(it)))
['int4', 'int5', 'int1', 'int2', 'int3']

Paulo Bu answer is the best pythonic way. If you want to stick with a function like yours:
def sort_array_by_second(a, b):
sorted = []
for n in b:
sorted.append(a[n-1])
return sorted
will do the trick.

Sorts A by the values of B:
A = ['int1', 'int2', 'int3', 'int4', 'int5']
B = [4, 5, 1, 2, 3]
from operator import itemgetter
C = [a for a, b in sorted(zip(A, B), key = itemgetter(1))]
print C
Output
['int3', 'int4', 'int5', 'int1', 'int2']

a = [11, 22, 44, 55] # values
b = [0, 1, 10, 11] # indexes to sort by
sorted_a = [-1] * (max(b) + 1)
for index, value in zip(b, a):
sorted_a[index] = value
print(sorted_a)
# -> [11, 22, -1, -1, -1, -1, -1, -1, -1, -1, 44, 55]

How to find the factors of a number and return it in a list in python [duplicate]

This question already has answers here:
What is the most efficient way of finding all the factors of a number in Python?
(29 answers)
Closed 9 years ago.
This is the code I have right now. I can't get it to return the right results for the question.
def problem(n):
myList = [1,n]
for i in range(1,n):
result = int(n ** .5)
new = n/result
i = i + 1
myList.append(new)
return myList

Factors of n are all numbers that divide into n evenly. So i is a factor of n if n % i == 0.
You need to do is perform this test for each number from 1 to n, and if that condition is true append that number to your list.
If you have issues as you start to write this code, update your question with what you tried.
Note that the above approach is not the most efficient way to find factors, but it seems to me like this is just an exercise for a beginning programmer so a naive approach is expected.

There are a few problems with your code. First of all you do not need to increment i as your for loop already does that. Secondly, using some basic math principles you only need to go through a range of numbers up to the square root of your passed in number. I will leave the second part for you to play and experiment with.
def problem(n):
myList = []
for i in range(1, n+1):
if n % i == 0:
myList.append(i)
return myList
For a more advanced approach you can try list comprehensions which are very powerful but are usually better for smaller data sets.
def problem(n):
return [x for x in range(1, n+1) if n % x == 0]

You only need to iterate from 1 to n ** 0.5 + 1, and your factors will be all i's, and n/i's you pick up along the way.
For example: factors of 10:
We only need to iterate from 1 to 4
i = 1 => 10 % 1 == 0, so factors: i = 1, 10 / i = 10
i = 2 => 10 % 2 == 0, so factors: i = 2, 10 / i = 5
i = 3 => 10 % 3 != 0, no factors
We don't need to go any further, the answer is 1, 2, 5, 10.
def problem(n):
myList = []
for i in xrange(1, int(n ** 0.5 + 1)):
if n % i == 0:
if (i != n/i):
myList.append(i)
myList.append(n / i)
else:
myList.append(i)
return myList
Result:
>>> problem(10)
[1, 10, 2, 5]
>>> problem(12)
[1, 12, 2, 6, 3, 4]
>>> problem(77)
[1, 77, 7, 11]
>>> problem(4)
[1, 4, 2]
>>> problem(64)
[1, 64, 2, 32, 4, 16, 8]
>>> len(problem(10 ** 12))
169

use a list comprehension:
In [4]: num=120
In [5]: [x for x in range(2,int(num/2)+1) if num%x==0]
Out[5]: [2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60]
In [6]: num=121
In [7]: [x for x in range(2,int(num/2)+1) if num%x==0]
Out[7]: [11]