Array min from argmin along an axis?
Hi folks, I need an efficient way to get both the min and argmin of a 2-d array along one axis. It seemed to me that the way to do this was to get the argmin and then use it to index into the array to get the min, but I can't figure out how to do it. Here's my toy example:
x = np.arange(25).reshape((5,5)) x array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24]]) y = np.abs(x - x.T) y array([[ 0, 4, 8, 12, 16], [ 4, 0, 4, 8, 12], [ 8, 4, 0, 4, 8], [12, 8, 4, 0, 4], [16, 12, 8, 4, 0]]) np.argmin(y, axis=0) array([0, 1, 2, 3, 4]) np.min(y, axis=0) array([0, 0, 0, 0, 0])
Here it seems like there should be a simple way to get the same array that min() returns using the argmin result, which won't need to 'search' in the array. Thanks very much, Ken
On Tue, Dec 13, 2011 at 4:11 PM, Ken Basye <kbasye1@jhu.edu> wrote:
Hi folks, I need an efficient way to get both the min and argmin of a 2-d array along one axis. It seemed to me that the way to do this was to get the argmin and then use it to index into the array to get the min, but I can't figure out how to do it. Here's my toy example:
x = np.arange(25).reshape((5,5)) x array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24]]) y = np.abs(x - x.T) y array([[ 0, 4, 8, 12, 16], [ 4, 0, 4, 8, 12], [ 8, 4, 0, 4, 8], [12, 8, 4, 0, 4], [16, 12, 8, 4, 0]]) np.argmin(y, axis=0) array([0, 1, 2, 3, 4]) np.min(y, axis=0) array([0, 0, 0, 0, 0])
Here it seems like there should be a simple way to get the same array that min() returns using the argmin result, which won't need to 'search' in the array.
You can use the result of argmin to index into y, if you combine it with, say, arange(ncols) in the second dimension: In [53]: y = random.randint(0,10,size=(5,7)) In [54]: y Out[54]: array([[3, 3, 5, 1, 5, 3, 7], [1, 0, 6, 8, 0, 1, 1], [7, 9, 9, 3, 3, 1, 6], [5, 3, 5, 4, 9, 7, 4], [1, 7, 1, 6, 6, 1, 8]]) In [55]: am = np.argmin(y, axis=0) In [56]: am Out[56]: array([1, 1, 4, 0, 1, 1, 1]) In [57]: colmins = y[am, arange(7)] In [58]: colmins Out[58]: array([1, 0, 1, 1, 0, 1, 1]) Warren
On Tue, Dec 13, 2011 at 22:11, Ken Basye <kbasye1@jhu.edu> wrote:
Hi folks, I need an efficient way to get both the min and argmin of a 2-d array along one axis. It seemed to me that the way to do this was to get the argmin and then use it to index into the array to get the min, but I can't figure out how to do it. Here's my toy example:
[~] |1> x = np.arange(25).reshape((5,5)) [~] |2> y = np.abs(x - x.T) [~] |3> y array([[ 0, 4, 8, 12, 16], [ 4, 0, 4, 8, 12], [ 8, 4, 0, 4, 8], [12, 8, 4, 0, 4], [16, 12, 8, 4, 0]]) [~] |4> i = np.argmin(y, axis=0) [~] |5> y[i, np.arange(y.shape[1])] array([0, 0, 0, 0, 0]) [~] |6> y[np.argmin(y, axis=0), np.arange(y.shape[1])] array([0, 0, 0, 0, 0]) -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco
|6> y[np.argmin(y, axis=0), np.arange(y.shape[1])] array([0, 0, 0, 0, 0])
Can xrange in this case save me from creating a temporary array here or doesn't it matter? |6> y[np.argmin(y, axis=0), xrange(y.shape[1])] array([0, 0, 0, 0, 0]) //Torgil On Tue, Dec 13, 2011 at 11:27 PM, Robert Kern <robert.kern@gmail.com> wrote:
On Tue, Dec 13, 2011 at 22:11, Ken Basye <kbasye1@jhu.edu> wrote:
Hi folks, I need an efficient way to get both the min and argmin of a 2-d array along one axis. It seemed to me that the way to do this was to get the argmin and then use it to index into the array to get the min, but I can't figure out how to do it. Here's my toy example:
[~] |1> x = np.arange(25).reshape((5,5))
[~] |2> y = np.abs(x - x.T)
[~] |3> y array([[ 0, 4, 8, 12, 16], [ 4, 0, 4, 8, 12], [ 8, 4, 0, 4, 8], [12, 8, 4, 0, 4], [16, 12, 8, 4, 0]])
[~] |4> i = np.argmin(y, axis=0)
[~] |5> y[i, np.arange(y.shape[1])] array([0, 0, 0, 0, 0])
[~] |6> y[np.argmin(y, axis=0), np.arange(y.shape[1])] array([0, 0, 0, 0, 0])
-- Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
participants (4)
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Ken Basye -
Robert Kern -
Torgil Svensson -
Warren Weckesser