There are two issues: the way you use the FFT, and the particular filter. Filtering is traditionally implemented as convolution in the time domain. You’re right that multiplying the spectra of the input and filter signals is equivalent. However, when you use the Discrete Fourier Transform (DFT) (implemented with a Fast Fourier Transform algorithm for … Read more
You can iterate that objects: function findObjectById($id){ $array = array( /* your array of objects */ ); foreach ( $array as $element ) { if ( $id == $element->id ) { return $element; } } return false; } Edit: Faster way is to have an array with keys equals to objects’ ids (if unique); Then … Read more
If you don’t care about creating a new dictionary with the desired items and throwing away the old one, simply try: dic = dic.Where(i => i.Value.BooleanProperty) .ToDictionary(i => i.Key, i => i.Value); If you can’t create a new dictionary and need to alter the old one for some reason (like when it’s externally referenced and … Read more
You can use GETFRAME function. It returns movie frame structure, which is actually rasterized figure. Field cdata will contain your matrix. F=getframe; figure(2) imagesc(F.cdata);
I think the two most straightforward approaches would be on click of any of the filter checkboxes either: Hide all <div> elements, then loop through the checkboxes and for each checked one .show() the <div> elements with the associated category. Loop through all checkboxes to make a list of the classes to be shown, then … Read more
I think you need boolean indexing: df1 = df[(df[‘category’] == ‘A’) & (df[‘value’].between(10,20))] print (df1) category value 2 A 15 4 A 18 And then: df2 = df[(df[‘category’] != ‘A’) & (df[‘value’].between(10,20))] print (df2) category value 1 B 10 Or: df3 = df[df[‘category’] != ‘A’] print (df3) category value 1 B 10 3 B 28 … Read more
The simplest way to format a cursor value is to use SimpleCursorAdapter.setViewBinder(..): SimpleCursorAdapter adapter = new SimpleCursorAdapter(this, R.layout.list, cursor, new String[] { Definition.Item.TITLE, Definition.Item.CREATE_DATE }, new int[] { R.id.title, R.id.createDate}); adapter.setViewBinder(new ViewBinder() { public boolean setViewValue(View aView, Cursor aCursor, int aColumnIndex) { if (aColumnIndex == 2) { String createDate = aCursor.getString(aColumnIndex); TextView textView = (TextView) … Read more
Yeah, a HashMap is not the right data structure for this. As Bozho said, a Trie would be the right one. With Java’s on-board tools, a TreeMap (or any SortedMap, actually) could be used: public <V> SortedMap<String, V> filterPrefix(SortedMap<String,V> baseMap, String prefix) { if(prefix.length() > 0) { char nextLetter = prefix.charAt(prefix.length() -1) + 1; String … Read more
How about a dict comprehension: filtered_dict = {k:v for k,v in d.iteritems() if filter_string in k} One you see it, it should be self-explanatory, as it reads like English pretty well. This syntax requires Python 2.7 or greater. In Python 3, there is only dict.items(), not iteritems() so you would use: filtered_dict = {k:v for … Read more
Here is some code I have written to do roughly what you describe. It creates a 3D data set (called cell, but it would be comparable to your array A), populates it with random data, and then computes a resultant 3D output array (called node, but it would be comparable to your array C) based … Read more