## Keep First Non-Zero Element, Set All Others to 0

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I have a 2-d NumPy array that looks like this:

array([[0. , 0. , 0.2, 0.2], [0.3, 0. , 0.3, 0. ]])

I'd like to modify it so that each row consists of all 0's, except for the first non-zero entry. If it's all 0s to start with, we don't change anything.

I could do this:

example = np.array([[0,0, 0.2, 0.2], [0.3, 0, 0.3, 0]]) my_copy = np.zeros_like(example) for i, row in enumerate(example): for j, elem in enumerate(row): if elem > 0: my_copy[i, j] = elem break

But that's ugly and not vectorized. Any suggestions for how to vectorize this?

Thanks!

Simply

e =np.zeros(example.shape) rows = np.arange(example.shape[0]) cols = np.argmax(example != 0, 1) e[rows, cols] = example[rows, cols]

**How to find first nonzero element/first '1' per row and set other ,** Learn more about indexing, matrix, first element, nonzero, set to zero, find, maximum. I've a Matrix f.ex. like. A=[0,1,1,0; 1,0,0,1;0,0,0,1;0,1,1,1]; A = 0 1 1 0. 1 0 0 1. 0 0 0 1. 0 1 1 1. only in 3D. I want to as a result only the first '1' of each row and all other elements or further "1" to zero. The result Get Support. Installation How to lookup first non-zero value and return corresponding column header in Excel? Supposing, you have a range of data, now, you want to return the column header in that row where the first non-zero value occurs as following screenshot shown, this article, I will introduce a useful formula for you to deal with this task in Excel.

Here's a vectorised solution. The trick is to calculate your first non-zero entries via `bool`

conversion and `argmax`

.

import numpy as np A = np.array([[0. , 0. , 0.2, 0.2], [0.3, 0. , 0.3, 0. ], [0. , 0. , 0. , 0. ]]) res = np.zeros(A.shape) idx = np.arange(res.shape[0]) args = A.astype(bool).argmax(1) res[idx, args] = A[idx, args] print(res) array([[ 0. , 0. , 0.2, 0. ], [ 0.3, 0. , 0. , 0. ], [ 0. , 0. , 0. , 0. ]])

**How to get the index of the first non-zero element in an 1-D array in ,** I want to find the index of the first non-zero element in a 1-D array in simulink, just like the way that 'find' command in matlab did. idx = find(x~=0, 1, 'first'); to zero. for example, if the eigenvalue small than 2, i set it to zero and keep those eigen Other MathWorks country sites are not optimized for visits from your location. extracting non-zero elements in a matrix or an Learn more about matrix, urgent, indexing MATLAB

*Setup*

x = np.array([[0. , 0. , 0.2, 0.2], [0.3, 0. , 0.3, 0. ], [0. , 0. , 0. , 0. ]])

Using ** logical_and** with

**:**

`np.eye`

m = (x!=0).argmax(1) x[~np.logical_and(x, np.eye(x.shape[1])[m])] = 0

Output:

array([[0. , 0. , 0.2, 0. ], [0.3, 0. , 0. , 0. ], [0. , 0. , 0. , 0. ]])

Using this method will be slightly slower than the other two suggested.

**Find indices and values of nonzero elements,** https://docs.scipy.org › doc › numpy › reference › generated › numpy.non I want to as a result only the first '1' of each row and all other elements or further "1" to zero. The result should be:

**numpy.nonzero,** numpy. nonzero (a)[source]¶. Return the indices of the elements that are non-zero. ..deprecated:: 1.17.0: Use atleast1d explicitly if this behavior is deliberate. I am doing the svd in Simulink, I got the eigenvalue, now I want to set some eigenvalues smaller than a given value to zero. for example, if the eigenvalue small than 2, i set it to zero and keep the those eigen values large than 2 unchanged, incuding the place in the matix and the value. PS: my matlab is 2009b, there are no Find block. so I

**Excel Formula To Get First Non-Zero Value in Row and Return ,** I accepted the answer above but wanted to show the other way I found to do it, in case it helps anyone. =OFFSET($I$8,0,MATCH(0,J10:N10,1)+1). You could also Re: How to globally change all z values to 0 The previous solution does not work for odd objects such as multilines. Here is a short lisp script which seems to work for most objects and allows the user to choose globally changing the Z coordinate or selecting a set to change.

**Nonzero Row,** A (0, 1)-valued matrix has the circular 1's property for columns if its rows can be First, that the nonzero rows, considered as vectors, form a linearly independent set, and The first nonzero element in the first nonzero row of a row-reduced matrix must be unity. Then, no other rows have a nonzero element in that column. I have an Na-by-Nt matrix which is sparse, i.e.: most elements are zeros. I want to find the indices of the non-zeros elements in the form of (i,j) where i is the row and j is the column.

##### Comments

- Please add a wanted result example. Thanks.
- Can you update yours so that it dynamically calculates rows, I don't want to mess up the answer :P I think it would just be
`np.arange(len(example))`

- Updated, very fast.
- Its faster with
`idx = np.arange(res.shape[0])`

for the first approach. - @Divakar yea the first and 3rd approaches are pretty much identical, but
`range`

is hella slow. Are you planning on putting us all to shame with an answer? :)