I have a matrix like following,
A =
1 2 3
4 5 6
7 8 9
10 11 12
4 5 6
7 8 9
4 5 6
1 2 3
I could extract unique rows in this matrix using command A_unique = unique(A,'rows') and result as follows
A_unique =
1 2 3
4 5 6
7 8 9
10 11 12
I need to find number of times each rows exists in the main matrix A
Some thing like following
A_unique_count =
2
3
2
1
How can I find count of unique rows? Can anyone help? Thanks in Advance
Manu
The third output of unique gives you the index of the unique row in the original array. You can use this with accumarray to count the number of occurrences.
For example:
A = [1 2 3
4 5 6
7 8 9
10 11 12
4 5 6
7 8 9
4 5 6
1 2 3];
[uniquerow, ~, rowidx] = unique(A, 'rows');
noccurrences = accumarray(rowidx, 1)
Returns:
noccurrences =
2
3
2
1
As expected
I would recommend #excaza's approach. But just for variety:
A_unique_count = diff([0; find([any(diff(sortrows(A), [], 1), 2); 1])]);
Related
I am trying to create two vectors
first:
[ 10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9 10 ]
second:
[ 1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1 ]
I had almost no problems with the first one:
i = 1:10;
for t = 1: 2*length(i)-1
y1(t) = abs(length(x)-t)+1;
end
But there are some problems with the second...
Does anyone have any idea how I can create it using the same for loop?Thanks in advance
If you want to do it with a loop:
N = 10;
for t = 1:2*N-1
y2(t) = -abs(t-N)+N;
end
But its probably easier to use the following, first make an array 1:N, and then concatenate the array N-1:-1:1:
y2 = [1:N, N-1:-1:1]
This question already has an answer here:
Pandas: how to create a simple counter that increases each n rows?
(1 answer)
Closed 1 year ago.
I have a dataframe which looks like this:
import pandas as pd
df = pd.DataFrame({
'SENDER_ID': [1,2,3,4,5,6,7,8,9,10,11,12] })
df =
SENDER_ID
0 1
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
Now I want to add a column which has the the element multiple times.
SENDER_ID counter
0 1 0
1 2 0
2 3 0
3 4 1
4 5 1
5 6 1
6 7 2
7 8 2
8 9 2
9 10 3
10 11 3
11 12 3
The dataframe always has a length of multiple of 3 and is much larger then in this simple example.
What is the easiest and most generic way to add this new column?
Another way using pd.RangeIndex:
df['count'] = pd.RangeIndex(0, len(df)//3).repeat(3)
print(df)
# Output:
SENDER_ID count
0 1 0
1 2 0
2 3 0
3 4 1
4 5 1
5 6 1
6 7 2
7 8 2
8 9 2
9 10 3
10 11 3
11 12 3
I think I found a solution which works:
max_list_length = int(len(df) / 3)
liste = [[n]*3 for n in range(0,max_list_length)]
value = sum(liste, [])
>>>> value
>>>> [0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3]
for n in range (0, len(df)):
df.at[n, 'counter'] = value[n]
I have a variable like this that is all one row:
1 2 3 4 5 6 7 8 9 2 4 5 6 5
I want to write a for loop that will find where a number is less than the previous one and put the rest of the numbers in a new row, like this
1 2 3 4 5 6 7 8 9
2 4 5 6
5
I have tried this:
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
m = zeros(size(test));
for i=1:numel(test)-1;
for rows=1:size(m,1)
if test(i) > test(i+1);
m(i+1, rows+1) = test(i+1:end)
end % for rows
end % for
But it's clearly not right and just hangs.
Let x be your data vector. What you want can be done quite simply as follows:
ind = [find(diff(x)<0) numel(x)]; %// find ends of increasing subsequences
ind(2:end) = diff(ind); %// compute lengths of those subsequences
y = mat2cell(x, 1, ind); %// split data vector according to those lenghts
This produces the desired result in cell array y. A cell array is used so that each "row" can have a different number of columns.
Example:
x = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
gives
y{1} =
1 2 3 4 5 6 7 8 9
y{2} =
2 4 5 6
y{3} =
5
If you are looking for a numeric array output, you would need to fill the "gaps" with something and filling with zeros seem like a good option as you seem to be doing in your code as well.
So, here's a bsxfun based approach to achieve the same -
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5] %// Input
idx = [find(diff(test)<0) numel(test)] %// positions of row shifts
lens = [idx(1) diff(idx)] %// lengths of each row in the proposed output
m = zeros(max(lens),numel(lens)) %// setup output matrix
m(bsxfun(#le,[1:max(lens)]',lens)) = test; %//'# put values from input array
m = m.' %//'# Output that is a transposed version after putting the values
Output -
m =
1 2 3 4 5 6 7 8 9
2 4 5 6 0 0 0 0 0
5 0 0 0 0 0 0 0 0
If an array contains N number of elements (elements can be repeated) and the goal is to make all the elements equal by a +1 on an element and a -1 on another element in each iteration, how can we determine whether it's possible or not to normalize the array? What will be the optimal algorithm to solve the problem?
Ex.
For the array 1 2 3, if I apply +1 on 1 and -1 on 3, the array becomes 2 2 2. That means it's possible in 1 iteration.
For the array 1 2 1, it's not possible to make all the elements equal.
First, since you're not disturbing the sum by each iteration, since you're increasing one number and decreasing another, the optimal target value is going to be the average.
If this average is a whole number, you should be able to achieve it with the iterations, however if the average is a fractional number then you will not be able to achieve it.
The number of steps is going to be the sum of the distances between each number and the target, divided by 2.
Every iteration pick one number above target and one below and apply the operations to them.
PS! As per commented, if all you want is answers to the following two questions:
Can it be done
What will the value be
Then the answers are:
Yes, provided the average number is a whole number
The value repeated in the whole array is the average number
Anyway, if you want the actual operations getting from the input to the target values, here's a longer example:
1 2 3 4 5 6 7 = 28, 28/7 = 4 (optimal target)
+ -
2 2 3 4 5 6 6
+ -
3 2 3 4 5 6 5
+ -
4 2 3 4 5 6 4
+ -
4 3 3 4 5 5 4
+ -
4 4 3 4 5 4 4
+ -
4 4 4 4 4 4 4
6 steps, let's total the distances from the first number:
1 2 3 4 5 6 7
3 2 1 0 1 2 3 = 12, divided by 2 = 6
Here's the example from the comments on the question:
1 9 10 12 3 7 = 42 / 6 = 7 (optimal target)
Distances:
1 9 10 12 3 7
6 2 3 5 4 0 = 20, divided by 2 = 10 (steps)
1 9 10 12 3 7
+ - step 1
2 8 10 12 3 7
+ - step 2
3 7 10 12 3 7
+ - step 3
4 7 9 12 3 7
+ - step 4
5 7 8 12 3 7
+ - step 5
6 7 7 12 3 7
+ - step 6
7 7 7 11 3 7
- + step 7
7 7 7 10 4 7
- + step 8
7 7 7 9 5 7
- + step 9
7 7 7 8 6 7
- + step 10
7 7 7 7 7 7
Here is a more pseudo-code like algorithm description:
Calculate SUM of all the elements
COUNT all the elements
If AVERAGE (SUM/COUNT) is not whole number, solution is not possible to achieve
STEPS = SUM(ABS(numberN - AVERAGE))/2
Each iteration, pick one number below AVERAGE and one above
Apply + operation to number below and - operation to number above
Repeat steps 5 and 6 until target achieved
Suppose I have a 2D array called A. I want to create a 3D array called B, whose "pages" are select columns of a stencil moving across A, column-by-column. For example, the first page of B might be the 1st, 3rd, and 5th columns of A. Then the second page would be the 2nd, 4th, and 6th columns of A, etc.
Anyone have an efficient way of doing this is MATLAB?
Thanks!
I am guessing you are looking for this -
%%// Given 2D array
A = randi(10,4,12)
t1 = reshape(A,size(A,1)*2,[]);
t2 = reshape(t1',size(A,2)/2,[],2); %%//'
B = permute(t2,[2 1 3]) %%// Output 3D array
Output -
A =
5 10 3 5 6 8 4 3 8 10 8 7
10 8 3 7 6 10 9 2 7 8 8 5
10 4 7 8 6 4 5 4 1 1 3 7
7 7 6 6 1 10 5 8 9 4 3 3
B(:,:,1) =
5 3 6 4 8 8
10 3 6 9 7 8
10 7 6 5 1 3
7 6 1 5 9 3
B(:,:,2) =
10 5 8 3 10 7
8 7 10 2 8 5
4 8 4 4 1 7
7 6 10 8 4 3
Of course, there is an alternative straight-forward approach for this special case -
B(:,:,1)=A(:,1:2:end);
B(:,:,2)=A(:,2:2:end);