Related
If there is an array that contains random integers in ascending order, how can I tell if this array contains a arithmetic sequence (length>3) with the common differece x?
Example:
Input: Array=[1,2,4,5,8,10,17,19,20,23,30,36,40,50]
x=10
Output: True
Explanation of the Example: the array contains [10,20,30,40,50], which is a arithmetic sequence (length=5) with the common differece 10.
Thanks!
I apologize that I have not try any code to solve this since I have no clue yet.
After reading the answers, I tried it in python.
Here are my codes:
df = [1,10,11,20,21,30,40]
i=0
common_differene=10
df_len=len(df)
for position_1 in range(df_len):
for position_2 in range(df_len):
if df[position_1] + common_differene == df[position_2]:
position_1=position_2
i=i+1
print(i)
However, it returns 9 instead of 4.
Is there anyway to prevent the repetitive counting in one sequence [10,20,30,40] and also prevent accumulating i from other sequences [1,11,21]?
You can solve your problem by using 2 loops, one to run through every element and the other one to check if the element is currentElement+x, if you find one that does, you can continue form there.
With the added rule of the sequence being more than 2 elements long, I have recreated your problem in FREE BASIC:
DIM array(13) As Integer = {1, 2, 4, 5, 8, 10, 17, 19, 20, 23, 30, 36, 40, 50}
DIM x as Integer = 10
DIM arithmeticArrayMinLength as Integer = 3
DIM index as Integer = 0
FOR position As Integer = LBound(array) To UBound(array)
FOR position2 As Integer = LBound(array) To UBound(array)
IF (array(position) + x = array(position2)) THEN
position = position2
index = index + 1
END IF
NEXT
NEXT
IF (index <= arithmeticArrayMinLength) THEN
PRINT false
ELSE
PRINT true
END IF
Hope it helps
Edit:
After reviewing your edit, I have come up with a solution in Python that returns all arithmetic sequences, keeping the order of the list:
def arithmeticSequence(A,n):
SubSequence=[]
ArithmeticSequences=[]
#Create array of pairs from array A
for index,item in enumerate(A[:-1]):
for index2,item2 in enumerate(A[index+1:]):
SubSequence.append([item,item2])
#finding arithmetic sequences
for index,pair in enumerate(SubSequence):
if (pair[1] - pair[0] == n):
found = [pair[0],pair[1]]
for index2,pair2 in enumerate(SubSequence[index+1:]):
if (pair2[0]==found[-1] and pair2[1]-pair2[0]==n):
found.append(pair2[1])
if (len(found)>2): ArithmeticSequences.append(found)
return ArithmeticSequences
df = [1,10,11,20,21,30,40]
common_differene=10
arseq=arithmeticSequence(df,common_differene)
print(arseq)
Output: [[1, 11, 21], [10, 20, 30, 40], [20, 30, 40]]
This is how you can get all the arithmetic sequences out of df for you to do whatever you want with them.
Now, if you want to remove the sub-sequences of already existing arithmetic sequences, you can try running it through:
def distinct(A):
DistinctArithmeticSequences = A
for index,item in enumerate(A):
for index2,item2 in enumerate([x for x in A if x != item]):
if (set(item2) <= set(item)):
DistinctArithmeticSequences.remove(item2)
return DistinctArithmeticSequences
darseq=distinct(arseq)
print(darseq)
Output: [[1, 11, 21], [10, 20, 30, 40]]
Note: Not gonna lie, this was fun figuring out!
Try from 1: check the presence of 11, 21, 31... (you can stop immediately)
Try from 2: check the presence of 12, 22, 32... (you can stop immediately)
Try from 4: check the presence of 14, 24, 34... (you can stop immediately)
...
Try from 10: check the presence of 20, 30, 40... (bingo !)
You can use linear searches, but for a large array, a hash map will be better. If you can stop as soon as you have found a sequence of length > 3, this procedure takes linear time.
Scan the list increasingly and for every element v, check if the element v + 10 is present and draw a link between them. This search can be done in linear time as a modified merge operation.
E.g. from 1, search 11; you can stop at 17; from 2, search 12; you can stop at 17; ... ; from 8, search 18; you can stop at 19...
Now you have a graph, the connected components of which form arithmetic sequences. You can traverse the array in search of a long sequence (or a longest), also in linear time.
In the given example, the only links are 10->-20->-30->-40->-50.
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
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.
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.
def unique(arr)
return arr.keep_if { |x| arr.count(x) == 1 }
end
print unique([2, 5, 5, 4, 22, 8, 2, 8])
#=> [4, 22, 2]
The value 2 appears twice in the array, but using the following method incorrectly returns it. Why does that happen, and what can I do to fix it?
Unfortunately, this is due to some hidden behavior in how keep_if works. To illustrate this behavior, we can make use of the lowest-hanging fruit in our debugging orchard, good ol' puts:
def unique(arr)
return arr.keep_if { |x|
puts x, arr.join(',')
arr.count(x) == 1
}
end
print unique([2, 5, 5, 4, 22, 8, 2, 8])
This gives us the following as output:
2
2,5,5,4,22,8,2,8
5
2,5,5,4,22,8,2,8
5
2,5,5,4,22,8,2,8
4
2,5,5,4,22,8,2,8
22
4,5,5,4,22,8,2,8
8
4,22,5,4,22,8,2,8
2
4,22,5,4,22,8,2,8
8
4,22,2,4,22,8,2,8
[4, 22, 2]
Look carefully at exactly what happens whenever the method discovers a new value it wants to keep: it stores that value in one of the early indexes in the array, overwriting what's already there. The next time it finds a value it wants to keep, it places it in the next spot, and so on.
This means that the first time keep_if looks at 2, it sees two of them and so decides to skip it. But it then sees a 4 that it wants to keep, and overwrites the first 2. Thus, the second time it sees a 2, it decides to keep it.
You are changing an array while you iterate over it. This is considered undefined behavior by Matz himself and you should avoid doing it if you want to avoid strange behavior like the one you see there.
Instead of your keep_if method, you should use this instead:
arr.select{ |x| arr.count(x) == 1 }
This won't change the array in place but will return a new one however.