This is a classic mistake. i in your case is already an element from array (i.e. another list), not an index of array (not an int), so if Volume == i[2]: counter += 1 You can check the Python tutorial. Also, try doing this: for i in array: print (i) And see what you get! … Read more
As I mentioned in the comments, it’s not practical to represent matrices using vector-of-vector for a few reasons: It is fiddly to set up; It is difficult to change; Cache locality is bad. Here is a very simple class I have created that will hold a 2D matrix in a single vector. This is pretty … Read more
The best way to do this would be to make b a set since you are only checking for membership inside it. >>> a = [1, 2, 3, 4, 5] >>> b = set([9, 7, 6, 5, 1, 0]) >>> [i for i, item in enumerate(a) if item in b] [0, 4]
The problem is: we have a std::tuple<T1, T2, …> and we have some function f that we can to call on each element, where f returns an int, and we want to store those results in an array. Let’s start with a concrete case: template <typename T> int f(T ) { return sizeof(T); } std::tuple<int, … Read more
s=”long string that I want to split up” indices = [0,5,12,17] parts = [s[i:j] for i,j in zip(indices, indices[1:]+[None])] returns [‘long ‘, ‘string ‘, ‘that ‘, ‘I want to split up’] which you can print using: print ‘\n’.join(parts) Another possibility (without copying indices) would be: s=”long string that I want to split up” indices = … Read more
Formally, all property names are strings. That means that array-like numeric property names really aren’t any different from any other property names. If you check step 6 in the relevant part of the spec, you’ll see that property accessor expressions are always coerced to strings before looking up the property. That process is followed (formally) … Read more
>>> a.argmax(axis=0) array([1, 1, 0])
dplyr has a group_indices() function that you can use like this: df %>% mutate(label = group_indices(., u, v)) %>% group_by(label) …
Unfortunately, lapply only gives you the elements of the vector you pass it. The usual work-around is to pass it the names or indices of the vector instead of the vector itself. But note that you can always pass in extra arguments to the function, so the following works: x <- list(a=11,b=12,c=13) # Changed to … Read more