I need to know, how to save integers from stdin into array, given by first integer in line... Ehm... hope you understand. I will give you an example.
On stdin I have:
0 : [ 1, 2, 3 ]
5 : [ 10, 11, 12, 13]
6 : [ 2, 4, 9 ]
0 : [ 4, 9, 8 ]
5 : [ 9, 6, 7 ]
5 : [ 1 ]
And I need save these integers to the arrays like this:
0={1, 2, 3, 4, 9, 8}
5={10, 11, 12, 13, 9, 6, 7, 1}
6={2, 4, 9}
I absolutely don't how to do it. There is a problem, that the number of arrays(in this case - 0, 5, 6 - so 3 arrays ) can be very high and I need to work effectively with memory...So I guess i will need something like malloc and free to solve this problem, or am I wrong? The names of arrays (0, 5, 6) can be changed. Number of integers in brackets has no maximum limit.
Thank you for any help.
I go with the assumption, this is homework, and I go with the assumption, this isn't your first homework to do, so I won't present you a solution but instead some tips that would help you to solve it yourself.
Given the input line
5 : [ 10, 11, 12, 13]
I will call "5" the "array name" and 10, 11, 12 and 13 the values to add.
You should implement some system to map array names to indices. A trivial approach would be like this:
.
size_t num_arrays;
size_t * array_names;
Here, in your example input, num_arrays will end up being 3 with array_names[3] = { 0, 5, 6}. If you find a new array name, realloc and add the new array name. Also you need the actual arrays for the values:
int * * array;
you need to realloc array for each new array name (like you realloc array_names). array[0] will represent array array_names[0] here array 0, array[1] will represent array array_names[1] here array 5 and array[2] will represent array array_names[2] here array 6.
To access an array, find it's index like so:
size_t index;
for (size_t index = 0; index < num_arrays && array_names[index] != search; ++index) ;
The second step is easy. Once you figured out, you need to use array[index] to add elemens, realloc that one (array[index] = realloc(array[index], new size)) and add elements there array[index][i+old_size] = new_value[i].
Obviously, you need to keep track of the number of elements in your separate arrays as well ;)
Hint: If searching for the array names take too long, you will have to replace that trivial mapping part by some more sophisticated data structure, like a hash map or a binary search tree. The rest of the concept may stay more or less the same.
Should you have problems to parse the input lines, I suggest, you open a new question specific on this parsing part.
In algorithmic terms, you need map (associative array) from ints to arrays. This is solved long ago in most higher level languages.
If you have to implement it manually, you have a few options:
simple "master" array where you store your 0, 5, 6, 1000000 and then map them to indices 0, 1, 2, 3 by doing search in for each time you have to access it (it's too time consuming when ;
hash table: write simple hash function to map 0, 5, 6, 1000000 (they're called keys) to values less than 1000, allocate array of 1000 elements and then make "master" array structures for each hash function result;
some kind of tree (e.g. red-black tree), may be a bit complex to implement manually.
Last two structures are part of programming classic and are well described in various articles and books.
Related
Say that I have a batch of arrays, and I would like to alter them based on conditions of particular values located by indices.
For example, say that I would like to increase and decrease particular values if the difference between those values are less than two.
For a single 1D array it can be done like this
import numpy as np
single2 = np.array([8, 8, 9, 10])
if abs(single2[1]-single2[2])<2:
single2[1] = single2[1] - 1
single2[2] = single2[2] + 1
single2
array([ 8, 7, 10, 10])
But I do not know how to do it for batch of arrays. This is my initial attempt
import numpy as np
single1 = np.array([6, 0, 3, 7])
single2 = np.array([8, 8, 9, 10])
single3 = np.array([2, 15, 15, 20])
batch = np.array([
np.copy(single1),
np.copy(single2),
np.copy(single3),
])
if abs(batch[:,1]-batch[:,2])<2:
batch[:,1] = batch[:,1] - 1
batch[:,2] = batch[:,2] + 1
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
Looking at np.any and np.all, they are used to create an array of booleans values, and I am not sure how they could be used in the code snippet above.
My second attempt uses np.where, using the method described here for comparing particular values of a batch of arrays by creating new versions of the arrays with values added to the front/back of the arrays.
https://stackoverflow.com/a/71297663/3259896
In the case of the example, I am comparing values that are right next to each other, so I created copies that shift the arrays forwards and backwards by 1. I also use only the particular slice of the array that I am comparing, since the other numbers would also be used in the comparison in np.where.
batch_ap = np.concatenate(
(batch[:, 1:2+1], np.repeat(-999, 3).reshape(3,1)),
axis=1
)
batch_pr = np.concatenate(
(np.repeat(-999, 3).reshape(3,1), batch[:, 1:2+1]),
axis=1
)
Finally, I do the comparisons, and adjust the values
batch[:, 1:2+1] = np.where(
abs(batch_ap[:,1:]-batch_ap[:,:-1])<2,
batch[:, 1:2+1]-1,
batch[:, 1:2+1]
)
batch[:, 1:2+1] = np.where(
abs(batch_pr[:,1:]-batch_pr[:,:-1])<2,
batch[:, 1:2+1]+1,
batch[:, 1:2+1]
)
print(batch)
[[ 6 0 3 7]
[ 8 7 10 10]
[ 2 14 16 20]]
Though I am not sure if this is the most computationally efficient nor programmatically elegant method for this task. Seems like a lot of operations and code for the task, but I do not have a strong enough mastery of numpy to be certain about this.
This works
mask = abs(batch[:,1]-batch[:,2])<2
batch[mask,1] -= 1
batch[mask,2] += 1
There two arrays, each of which will always contain an even, (though not equal) number of integers so that each pair will form a range, eg. 1..5, 8..12, etc.
var defaultArray: [Int] = [1, 5, 8, 12]
var priorityArray: [Int] = [1, 3, 5, 10, 13, 20]
What I'm looking for is a generic algorithm that will find each occurrence of where a range from priorityArray overlaps a range from defaultArray and will insert the priorityRange into the defaultArray while splitting the defaultRange apart if necessary.
The goal is to have a combined array of ranges while maintaining their original "types" like so:
var result: [Int] = [
1, 3, // priority
3, 5, // default
5, 10, // priority
10, 12, // default
13, 20 // priority
]
I'll use a simple struct to illustrate the final desired result:
var result: [Range] = [
Range(from: 1, until: 3, key: "priority"),
Range(from: 3, until: 5, key: "default"),
Range(from: 5, until: 10, key: "priority"),
Range(from: 10, until: 12, key: "default"),
Range(from: 13, until: 20, key: "priority")
]
We start with those arrays def and prio and first check if the intervals themselves are sorted wrt their start/end points. Those arrays would then contain the smallest number in the first array position. Ensure these arrays are simple/correct (=no overlapping intervals). If they are not, you can simplify/sanitise them.
We then initialise
array index d=0 to index the def array
array index p=0 to index the prio array
a new array result to hold all your newly created intervals.
a variable s=none to hold the current status
We now determine if the relation between the def[d] and prio[p].
If def[d]<prio[p], we set t=def[d], increment d and set s=def.
If def[d]> prio[p], we set t=prio[p], increment p and set s=prio.
If they are equal, we set t=prio[p], increment p and d and set s=both.
We can now initialise a new entry for the result array with start=def[0]. The priority is either def (if s==def) or prio (if s was prio or both). To determine the end, you can again compare def[d] with prio[p] to determine where it should end. At this point, you should adjust s again, but ensure that you keep track of the proper state which you're in (going from both to def, prio or none depending on the relation between def[d] and prio[p]). As mentioned in the comments of the OP, the different possibilities might require more clarification, but you should be able to incorporate them into a state.
Going from there, you can keep iterating and adjusting your variables until both are done (with d=len(def) and p=len(prio). You should end up with a nice array containing all the desired consolidated intervals.
This is basically a stateful sweep through the 2 arrays, keeping track of the current position in the integer range and advancing 1 (maybe 2) position(s) at a time.
If I have the array as [7, 11, 13, 9, 4, 6], and the user input is 10
I want the integer just above (or equal to ) 10 (in this case 11) and replace it with integer - 10 (11 - 10)
I can't order the list and binary search as I have to return the index of updated element.
I looked into ordering the (index, integer) pairs and then binary searching but then after updating I have to re sort the array of pairs for next user input.
original array = [7, 11, 13, 9, 4, 6]
sorted array = [4, 6, 7, 9, 11, 13]
user input = 10
output = 1 (index of updated integer)
updated sorted array = [4, 6, 7, 9, 1, 13]
re sorted array = [1, 4, 6, 7, 9, 13]
user input = 4
...
What data structure would be suitable to implement this?
You can use an ordered set to store (value, index) tuples. You can define custom comparison functions to fit your goal (it depends on the programming language in question).
Ordered set operations include:
Add an element.
Remove an element.
Search for an element.
Search for an element greater or equal than given element.
Search for an element greater than given element.
The given problem can be solved using an ordered set with operations as described above.
Some programming languages (like set in c++) have predefined data structures so you don't need to implement it from scratch (only to use it). Ordered set internal implementation usually is Red-Black Tree (c++) or AVL Tree.
Check if your programming language has a built-in ordered set or use some library that include it. It's a very useful and common data structure so it shouldn't be hard to find.
Im trying to manual sort on the below array.
The issue here is, the result varies while reading the item from the "for-loop enumuration" (noted as //(2)) verses reading it as a subscript (noted as //(1)). It could be a minor issue hiding behind my eye. Appreciate your time.
var mySortArray : Array<Int> = []
mySortArray = [1,5,3,3,21,11,2]
for (itemX,X) in mySortArray.enumerated() {
for (itemY,Y) in mySortArray.enumerated() {
// if mySortArray[itemX] < mySortArray[itemY] // (1)
if X < Y // (2)
{
//Swap the position of item in the array
mySortArray.swapAt(itemX, itemY)
}
}
}
print(mySortArray)
// Prints [1, 2, 3, 3, 5, 11, 21] ( for condition // (1))
// Prints [2, 1, 3, 5, 11, 3, 21] ( for condition // (2))
mySortArray = [1,5,3,3,21,11,2]
print("Actual Sort Order : \(mySortArray.sorted())")
// Prints Actual Sort Order : [1, 2, 3, 3, 5, 11, 21]
The problem here is that the function .enumerated() returns a new sequence and iterates that. Think of it as a new array.
So, you are working with 3 different arrays here.
You have an unsorted array that you want to fix. Lets call this the w ("working array") and then you have you array x and array y.
So, w is [1,5,3,3,21,11,2], x and y are effectively the same as w at the beginning.
Now you get your first two values that need to swap...
valueX is at index 1 of x (5). valueY is at index 2 of y (3).
And you swap them... in w.
So now w is [1,3,5,3,21,11,2] but x and y are unchanged.
So now you indexes are being thrown off. You are comparing items in x with items in y and then swapping them in we which is completely different.
You need to work with one array the whole time.
Of course... there is also the issue that your function is currently very slow. O(n^2) and there are much more efficient ways of sorting.
If you are doing this as an exercise in learning how to write sort algorithms then keep going. If not you should really be using the .sort() function.
Really what you want to be doing is not using .enumerated() at all. Just use ints to get (and swap) values in w.
i.e. something like
for indexX in 0..<w.count {
for indexY in indexX..<w.count {
// do some comparison stuff.
// do some swapping stuff.
}
}
If I have the following array:
x = double([1, 1, 1, 10, 1, 1, 50, 1, 1, 1 ])
I want to do the following:
Group the array into groups of 5 which will each be evaluated separately.
Identify the MAX value each of the groups of the array
Remove that MAX value and put it into another array.
Finally, I want to print the updated array x without the MAX values, and the new array containing the MAX values.
How can I do this? I am new to IDL and have had no formal training in coding.
I understand that I can write the code to group and find the max values this way:
FOR i = 1, (n_elements(x)-4) do begin
print, "MAX of array", MAX( MAX(x[i-1:1+3])
ENDFOR
However, how do I implement all of what I specified above? I know I have to create an empty array that will append the values found by the for loop, but I don't know how to do that.
Thanks
I changed your x to have unique elements to make sure I wasn't fooling myself. It this, the number of elements of x must be divisible by group_size:
x = double([1, 2, 3, 10, 4, 5, 50, 6, 7, 8])
group_size = 5
maxes = max(reform(x, group_size, n_elements(x) / group_size), ind, dimension=1)
all = bytarr(n_elements(x))
all[ind] = 1
x_without_maxes = x[where(all eq 0)]
print, maxes
print, x_without_maxes
Lists are good for this, because they allow you to pop out values at specific indices, rather than rewriting the whole array again. You might try something like the following. I've used a while loop here, rather than a for loop, because it makes it a little easier in this case.
x = List(1, 1, 1, 10, 1, 1, 50, 1, 1, 1)
maxValues = List()
pos = 4
while (pos le x.length) do begin
maxValues.add, max(x[pos-4:pos].toArray(), iMax)
x.Remove, iMax+pos-4
pos += 5-1
endwhile
print, "Max Values : ", maxValues.toArray()
print, "Remaining Values : ", x.toArray()
This allows you to do what you want I think. At the end, you have a List object (which can easily be converted to an array) with the max values for each group of 5, and another containing the remaining values.
Also, please tag this as idl-programming-language rather than idl. They are two different tags.