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import pandas as pd
import numpy as np
column = np.array([5505, 5505, 5505, 34565, 34565, 65539, 65539])
column = pd.Series(column)
myDict = column.groupby(by = column ).groups
I am creating a dictionary from a pandas df using df.group(by=..) which has the form:
>>> myDict
{5505: Int64Index([0, 1, 2], dtype='int64'), 65539: Int64Index([5, 6], dtype='int64'), 34565: Int64Index([3, 4], dtype='int64')}
I have a numpy array, e.g.
myArray = np.array([34565, 34565, 5505,65539])
and I want to replace each of the array's elements with the dictionary's values.
I have tried several solutions that I have found (e.g. here and here) but these examples have dictionaries with single dictionary values, and I am always getting the error of setting an array element with a sequence. How can I get over this problem?
My intended output is
np.array([3, 4, 3, 4, 0, 1, 2, 5, 6])
One approach based on np.searchsorted -
# Extract dict info
k = list(myDict.keys())
v = list(myDict.values())
# Use argsort of k to find search sorted indices from myArray in keys
# Index into the values of dict based on those indices for output
sidx = np.argsort(k)
idx = sidx[np.searchsorted(k,myArray,sorter=sidx)]
out_arr = np.concatenate([v[i] for i in idx])
Sample input, output -
In [369]: myDict
Out[369]:
{5505: Int64Index([0, 1, 2], dtype='int64'),
34565: Int64Index([3, 4], dtype='int64'),
65539: Int64Index([5, 6], dtype='int64')}
In [370]: myArray
Out[370]: array([34565, 34565, 5505, 65539])
In [371]: out_arr
Out[371]: array([3, 4, 3, 4, 0, 1, 2, 5, 6])
This question already has answers here:
In Swift, an efficient function that separates an array into 2 arrays based on a predicate
(7 answers)
Closed 6 months ago.
Problem
Given an array of values how can I split it into sub-arrays made of elements that are equal?
Example
Given this array
let numbers = [1, 1, 1, 3, 3, 4]
I want this output
[[1,1,1], [3, 3], [4]]
What I am NOT looking for
A possible way of solving this would be creating some sort of index to indicate the occurrences of each element like this.
let indexes = [1:3, 3:2, 4:1]
And finally use the index to rebuild the output array.
let subsequences = indexes.sort { $0.0.0 < $0.1.0 }.reduce([Int]()) { (res, elm) -> [Int] in
return res + [Int](count: elm.1, repeatedValue: elm.0)
}
However with this solution I am losing the original values. Of course in this case it's not a big problem (an Int value is still and Inteven if recreated) but I would like to apply this solution to more complex data structures like this
struct Starship: Equatable {
let name: String
let warpSpeed: Int
}
func ==(left:Starship, right:Starship) -> Bool {
return left.warpSpeed == right.warpSpeed
}
Final considerations
The function I am looking for would be some kind of reverse of flatten(), infact
let subsequences: [[Int]] = [[1,1,1], [3, 3], [4]]
print(Array(subsequences.flatten())) // [1, 1, 1, 3, 3, 4]
I hope I made myself clear, let me know should you need further details.
// extract unique numbers using a set, then
// map sub-arrays of the original arrays with a filter on each distinct number
let numbers = [1, 1, 1, 3, 3, 4]
let numberGroups = Set(numbers).map{ value in return numbers.filter{$0==value} }
print(numberGroups)
[EDIT] changed to use Set Initializer as suggested by Hamish
[EDIT2] Swift 4 added an initializer to Dictionary that will do this more efficiently:
let numberGroups = Array(Dictionary(grouping:numbers){$0}.values)
For a list of objects to be grouped by one of their properties:
let objectGroups = Array(Dictionary(grouping:objects){$0.property}.values)
If you could use CocoaPods/Carthage/Swift Package Manager/etc. you could use packages like oisdk/SwiftSequence which provides the group() method:
numbers.lazy.group()
// should return a sequence that generates [1, 1, 1], [3, 3], [4].
or UsrNameu1/TraverSwift which provides groupBy:
groupBy(SequenceOf(numbers), ==)
If you don't want to add external dependencies, you could always write an algorithm like:
func group<S: SequenceType where S.Generator.Element: Equatable>(seq: S) -> [[S.Generator.Element]] {
var result: [[S.Generator.Element]] = []
var current: [S.Generator.Element] = []
for element in seq {
if current.isEmpty || element == current[0] {
current.append(element)
} else {
result.append(current)
current = [element]
}
}
result.append(current)
return result
}
group(numbers)
// returns [[1, 1, 1], [3, 3], [4]].
Let's assume that you have an unsorted array of items. You will need to sort the initial array then you will have something like this:
[1, 1, 1, 3, 3, 4]
After that you will initialize two arrays: one for storing arrays and another one to use it as a current array.
Loop through the initial array and:
if the current value isn't different from the last one, push it to the current array
otherwise push the current array to the first one then empty the current array.
Hope it helps!
Worth mentioning, using Swift Algorithms this is now a one-liner:
import Algorithms
let numbers = [1, 1, 1, 3, 3, 4]
let chunks: [[Int]] = numbers.chunked(by: ==).map { .init($0) }
print(chunks) // [[1, 1, 1], [3, 3], [4]]
I would like to turn an array into an array of arrays following another array of arrays. I'm not sure how to do this, here are the arrays:
orig_array = [[0,1],[4],[3],[],[3,2,6],[]]
my_array = [2,0,1,3,3,4,5]
wanted_array = [[2,0],[1],[3],[],[3,4,5],[]]
I would like to keep the empty arrays.
Thanks
Get the lengths of each element in orig_array, perform cumumlative summations along the length values to give us the indices at which my_array needs to be split and finally use np.split to actually perform the splitting. Thus, the implementation would look something like this -
lens = [len(item) for item in orig_array]
out = np.split(my_array,np.cumsum(lens))[:-1]
Sample run -
In [72]: orig_array = np.array([[0,1],[4],[3],[],[3,2,6],[]])
...: my_array = np.array([2,0,1,3,3,4,5])
...:
In [73]: lens = [len(item) for item in orig_array]
...: out = np.split(my_array,np.cumsum(lens))[:-1]
...:
In [74]: out
Out[74]:
[array([2, 0]),
array([1]),
array([3]),
array([], dtype=int64),
array([3, 4, 5]),
array([], dtype=int64)]
def do(format, values):
if type(format) == list:
return [do(v, values) for v in format]
else:
return values.pop(0)
print do(orig_array, my_array)
Note: this destroys the array where the values come from.
You could do the following:
import copy
def reflect_array(orig_array, order):
wanted_array = copy.deepcopy(orig_array)
for i, part_list in enumerate(orig_array):
for j, _ in enumerate(part_list):
wanted_array[i][j] = order.pop()
return wanted_array
Test run:
orig_array = [[0,1],[4],[3],[],[3,2,6],[]]
my_array = [2,0,1,3,3,4,5]
print reflect_array(orig_array, my_array)
# [[2, 0], [1], [3], [], [3, 4, 5], []]
In [858]: my_array = [2,0,1,3,3,4,5]
In [859]: [[my_array.pop(0) for _ in range(len(x))] for x in orig_array]
Out[859]: [[2, 0], [1], [3], [], [3, 4, 5], []]
Use b=my_array[:] if you don't want to change my_array.
This operates on the same principle as #karoly's answer; just more direct because it assumes only one level of nesting.
Is there a sensible way to do the following:
I want to take an array and select specific items from the array according to conditions, removing them from the array as they go.
(I basically want to split the contents of an array into categories).
array = [1,2,3,4,5,6,7,8]
less_than_three = array.reject_destructively{|v| v<3}
=> [1,2]
array
=> [3,4,5,6,7,8]
more_than_five = array.reject_destructively{|v| v>5}
=> [6,7,8]
array
=> [3,4,5]
I've tried delete_if, select!, reject! and none of them seem to be able to give you the affected items whilst leaving the array with the rest.
Unless I'm going mad, which is entirely possible.
As I understood the question, you do not want to produce two new objects. Here you go:
class Array
def carve!
dup.tap { delete_if &Proc.new } - self
end
end
array = [1,2,3,4,5,6,7,8]
p array.carve! { |v| v < 3 }
#⇒ [1, 2] # returned by Array#carve method
p array
#⇒ [3, 4, 5, 6, 7, 8] # remained in original array
Using this solution, array.__id__ remains the same. And this is the golfiest answer all around :)
You can build your own method for this...
class Array
def extract(&block)
temp = self.select(&block)
self.reject!(&block)
temp
end
end
then...
a = [1, 2, 3, 4, 5]
a.extract{|x| x < 3}
=> [1,2]
p a
=> [3, 4, 5]
EDIT: If you don't want to monkey patch (but monkey patching isn't evil in itself) you can do it with a vanilla method...
def select_from_array(array, &block)
temp = array.select(&block)
array.reject!(&block)
temp
end
array = [1,2,3,4,5,6,7,8]
less_than_three = select_from_array(array){|v| v<3}
=> [1,2]
array
=> [3,4,5,6,7,8]
more_than_five = select_from_array(array){|v| v>5}
=> [6,7,8]
array
=> [3,4,5]
In rails 6 there is a method extract!:
a = [1, 2, 3] #=> [1, 2, 3]
a.extract! { |num| num.odd? } #=> [1, 3]
a #=> [2]
irb(main):001:0> array = [1,2,3,4,5,6,7,8]
=> [1, 2, 3, 4, 5, 6, 7, 8]
irb(main):002:0> array.partition{|v| v < 3}
=> [[1, 2], [3, 4, 5, 6, 7, 8]]
is there a specific reason, why this has to be destructive ?
Will this help
class Array
def reject_destructively(&block)
arr = self.select(&block)
arr.each{ |i| self.delete(i) }
arr
end
end
array = [1,2,3,4,5,6,7,8]
p less_than_three = array.reject_destructively{|v| v<3}
#=> [1,2]
p array
#=> [3,4,5,6,7,8]
p more_than_five = array.reject_destructively{|v| v>5}
#=> [6,7,8]
p array
#=> [3,4,5]
The above code can be simplified further to look like:
class Array
def reject_destructively(&block)
self.select(&block).each{ |i| self.delete(i) }
end
end
Ok. This works, avoids monkey patching, keeps it to one line...etc, but it's damn ugly....
less_than_three = array.dup - array.reject!{|v| v<3}
=> [1,2]
array
=> [3,4,5,6,7,8]
more_than_five = array.dup - array.reject!{|v| v>5}
=> [6,7,8]
array
=> [3,4,5]
module Enumerable
def reject_destructively
array=[]
self.each do |y|
if yield(y)
array<<y
end
end
array.each do |x|
self.delete(x)
end
return array
end
end
array=[10,9,2,1,3,45,52]
print less_than_three = array.reject_destructively{|v| v < 3}
print array
You can use group_by to get all of the elements that satisfy the condition in one group, and all of the rest in the other.
For example
[1,2,3,4,5].group_by{|i| i > 3}
gives
{false=>[1, 2, 3], true=>[4, 5]}
More information is available at http://ruby-doc.org/core-2.1.1/Enumerable.html#method-i-group_by
Let's say I am trying to remove elements from array a = [1,1,1,2,2,3]. If I perform the following:
b = a - [1,3]
Then I will get:
b = [2,2]
However, I want the result to be
b = [1,1,2,2]
i.e. I only remove one instance of each element in the subtracted vector not all cases. Is there a simple way in Ruby to do this?
You may do:
a= [1,1,1,2,2,3]
delete_list = [1,3]
delete_list.each do |del|
a.delete_at(a.index(del))
end
result : [1, 1, 2, 2]
[1,3].inject([1,1,1,2,2,3]) do |memo,element|
memo.tap do |memo|
i = memo.find_index(e)
memo.delete_at(i) if i
end
end
Not very simple but:
a = [1,1,1,2,2,3]
b = a.group_by {|n| n}.each {|k,v| v.pop [1,3].count(k)}.values.flatten
=> [1, 1, 2, 2]
Also handles the case for multiples in the 'subtrahend':
a = [1,1,1,2,2,3]
b = a.group_by {|n| n}.each {|k,v| v.pop [1,1,3].count(k)}.values.flatten
=> [1, 2, 2]
EDIT: this is more an enhancement combining Norm212 and my answer to make a "functional" solution.
b = [1,1,3].each.with_object( a ) { |del| a.delete_at( a.index( del ) ) }
Put it in a lambda if needed:
subtract = lambda do |minuend, subtrahend|
subtrahend.each.with_object( minuend ) { |del| minuend.delete_at( minuend.index( del ) ) }
end
then:
subtract.call a, [1,1,3]
A simple solution I frequently use:
arr = ['remove me',3,4,2,45]
arr[1..-1]
=> [3,4,2,45]
a = [1,1,1,2,2,3]
a.slice!(0) # remove first index
a.slice!(-1) # remove last index
# a = [1,1,2,2] as desired
For speed, I would do the following, which requires only one pass through each of the two arrays. This method preserves order. I will first present code that does not mutate the original array, then show how it can be easily modified to mutate.
arr = [1,1,1,2,2,3,1]
removals = [1,3,1]
h = removals.group_by(&:itself).transform_values(&:size)
#=> {1=>2, 3=>1}
arr.each_with_object([]) { |n,a|
h.key?(n) && h[n] > 0 ? (h[n] -= 1) : a << n }
#=> [1, 2, 2, 1]
arr
#=> [1, 1, 1, 2, 2, 3, 1]
To mutate arr write:
h = removals.group_by(&:itself).transform_values(&:count)
arr.replace(arr.each_with_object([]) { |n,a|
h.key?(n) && h[n] > 0 ? (h[n] -= 1) : a << n })
#=> [1, 2, 2, 1]
arr
#=> [1, 2, 2, 1]
This uses the 21st century method Hash#transform_values (new in MRI v2.4), but one could instead write:
h = Hash[removals.group_by(&:itself).map { |k,v| [k,v.size] }]
or
h = removals.each_with_object(Hash.new(0)) { | n,h| h[n] += 1 }