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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.
I'm working in a PSET6 for CS50 on edx, it's called DNA
https://cs50.harvard.edu/x/2020/psets/6/dna/
And a video to explain the problem in details : https://youtu.be/j84b_EgntcQ
Here's the code down below that I need help with
I want to count how many times does the following set of characters repeats consecutively
"AGATC", "AATG", and "TATC"
which means if "AGATC" appeared one time , I ignore it,
however if it's repeated back to back , so I count those, and so on , then return the maximum number was it counted
Here's a text for , you are free to edit for testing
that code , doesn't provide the needed results, because counter is grouping each letter
so is there's a way i can get the below result
# code from https://www.journaldev.com/23666/python-string-find
def find_all_indexes(input_str, search_str):
l1 = []
length = len(input_str)
index = 0
while index < length:
i = input_str.find(search_str, index)
if i == -1:
return l1
l1.append(i)
index = i + 1
return l1
s = 'AAGAGATCAGATCAGATCAGATCAGGTGAGTTAAATAGAAGATCAGATCAGATCAGATCAGATCATAGGTTAAAAATGAATGAATGAATGAATGATTAAAGGAGATCAGATCAGATCAGATCTATCTATCTATCTATCAATGAATGAATGTATCTATCTATCAGAAAATGAATGAATGAAGAGTATATCTATCAATAGTTAAAGAGTAAGATATTGAATTGAAAATATTGTTGGGGAAAGGAGGGATAGAAGG'
print(find_all_indexes(s, 'AGATC'))
# printed values: [3, 8, 13, 18, 39, 44, 49, 54, 59, 102, 107, 112, 117]
so now I'm able to find the location
however i don't know how to count each consecutive ones
for example
locations 3 , 8 , 13 , 39, 44,54 are counted 6
then locations 102, 107, 112, 117 are counted 4
so the greatest number is 6
then I need to get this 6 please which is the maximum repeats this string was repeated
The code is based on your example, maybe you would need to optimize it:
import re
def find_all_indexes(input_str, search_str):
max_count = 0
for group in re.findall(f"(?:{search_str})+", input_str):
count = len(re.findall(f"(?:{search_str})", group))
if count > max_count:
max_count = count
return max_count
s = 'AAGAGATCAGATCAGATCAGATCAGGTGAGTTAAATAGAAGATCAGATCAGATCAGATCAGATCATAGGTTAAAAATGAATGAATGAATGAATGATTAAAGGAGATCAGATCAGATCAGATCTATCTATCTATCTATCAATGAATGAATGTATCTATCTATCAGAAAATGAATGAATGAAGAGTATATCTATCAATAGTTAAAGAGTAAGATATTGAATTGAAAATATTGTTGGGGAAAGGAGGGATAGAAGG'
print(find_all_indexes(s, 'AGATC'))
Take a look again. You can check the documentation here: https://docs.python.org/3/library/re.html#re.findall
Let's assume a multi-dimensional array
import numpy as np
foo = np.random.rand(102,43,35,51)
I know that those last dimensions represent a 2D space (35,51) of which I would like to index a range of rows of a column
Let's say I want to have rows 8 to 30 of column 0
From my understanding of indexing I should call
foo[0][0][8::30][0]
Knowing my data though (unlike the random data used here), this is not what I expected
I could try this that does work but looks ridiculous
foo[0][0][[8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30],0]
Now from what I can find in this documentation I can also use
something like:
foo[0][0][[8,30],0]
which only gives me the values of rows 8 and 30
while this:
foo[0][0][[8::30],0]
gives an error
File "<ipython-input-568-cc49fe1424d1>", line 1
foo[0][0][[8::30],0]
^
SyntaxError: invalid syntax
I don't understand why the :: argument cannot be passed here. What is then a way to indicate a range in your indexing syntax?
So I guess my overall question is what would be the proper pythonic equivalent of this syntax:
foo[0][0][[8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30],0]
Instead of
foo[0][0][8::30][0]
try
foo[0, 0, 8:30, 0]
The foo[0][0] part is the same as foo[0, 0, :, :], selecting a 2d array (35 x 51). But foo[0][0][8::30] selects a subset of those rows
Consider what happens when is use 0::30 on 2d array:
In [490]: np.zeros((35,51))[0::30].shape
Out[490]: (2, 51)
In [491]: np.arange(35)[0::30]
Out[491]: array([ 0, 30])
The 30 is the step, not the stop value of the slice.
the last [0] then picks the first of those rows. The end result is the same as foo[0,0,0,:].
It is better, in most cases, to index multiple dimensions with the comma syntax. And if you want the first 30 rows use 0:30, not 0::30 (that's basic slicing notation, applicable to lists as well as arrays).
As for:
foo[0][0][[8::30],0]
simplify it to x[[8::30], 0]. The Python interpreter accepts [1:2:3, 0], translating it to tuple(slice(1,2,3), 0) and passing it to a __getitem__ method. But the colon syntax is accepted in a very specific context. The interpreter is treating that inner set of brackets as a list, and colons are not accepted there.
foo[0,0,[1,2,3],0]
is ok, because the inner brackets are a list, and the numpy getitem can handle those.
numpy has a tool for converting a slice notation into a list of numbers. Play with that if it is still confusing:
In [495]: np.r_[8:30]
Out[495]:
array([ 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29])
In [496]: np.r_[8::30]
Out[496]: array([0])
In [497]: np.r_[8:30:2]
Out[497]: array([ 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28])
I have a multiple input arrays and I want to generate one output array where the value is 0 if all elements in a column are the same and the value is 1 if all elements in a column are different.
For example, if there are three arrays :
A = [28, 28, 43, 43]
B = [28, 43, 43, 28]
C = [28, 28, 43, 43]
Output = [0, 1, 0, 1]
The arrays can be of any size and any number, but the arrays are also the same size.
A none loopy way is to use diff and any to advantage:
A = [28, 28, 43,43];
B = [28, 43, 43,28];
C = [28, 28, 43,43];
D = any(diff([A;B;C])) %Combine all three (or all N) vectors into a matrix. Using the Diff to find the difference between each element from row to row. If any of them is non-zero, then return 1, else return 0.
D = 0 1 0 1
There are several easy ways to do it.
Let's start by putting the relevant vectors in a matrix:
M = [A; B; C];
Now we can do things like:
idx = min(M)==max(M);
or
idx = ~var(M);
No one seems to have addressed that you have a variable amount of arrays. In your case, you have three in your example but you said you could have a variable amount. I'd also like to take a stab at this using broadcasting.
You can create a function that will take a variable number of arrays, and the output will give you an array of an equal number of columns shared among all arrays that conform to the output you're speaking of.
First create a larger matrix that concatenates all of the arrays together, then use bsxfun to take advantage of broadcasting the first row and ensuring that you find columns that are all equal. You can use all to complete this step:
function out = array_compare(varargin)
matrix = vertcat(varargin{:});
out = ~all(bsxfun(#eq, matrix(1,:), matrix), 1);
end
This will take the first row of the stacked matrix and see if this row is the same among all of the rows in the stacked matrix for every column and returns a corresponding vector where 0 denotes each column being all equal and 1 otherwise.
Save this function in MATLAB and call it array_compare.m, then you can call it in MATLAB like so:
A = [28, 28, 43, 43];
B = [28, 43, 43, 28];
C = [28, 28, 43, 43];
Output = array_compare(A, B, C);
We get in MATLAB:
>> Output
Output =
0 1 0 1
Not fancy but will do the trick
Output=nan(length(A),1); %preallocation and check if an index isn't reached
for i=1:length(A)
Output(i)= ~isequal(A(i),B(i),C(i));
end
If someone has an answer without the loop take that, but i feel like performance is not an issue here.
I have an array that is already sorted in partitions of 4:
2, 23, 45, 55, 1, 4, 23, 74545, 75, 234, 323, 9090, 2, 43, 6342, 323452
What would be the most efficient way to sort this array? Note: the array size is always even and the program knows that every 4 elements are sorted.
I think you can use merge sort for problems like this.
You might be able to use strand sort for this.