Here is a more efficient algorithm: For each unique number that is present in at least one of the sublists, let’s maintain a list of indices of all sublists that contain this number. This part is O(n * log n) time if we use sorting to find unique numbers or O(n) if we use a … Read more
Perhaps the closest thing in Python to this Javascript array behavior is a dictionary. It’s a hashed mapping. defaultdict is a dictionary that implements a default value. In [1]: from collections import defaultdict In [2]: arr = defaultdict(bool) Insert a couple of True elements: In [3]: arr[10] = True In [4]: arr Out[4]: defaultdict(bool, {10: … Read more
I am note sure if you can omit that loop in an efficient way, but when we do not use np.delete() but just use the indices to shrink payoff in every iteration I already got a big speed increase on my machine: # … arr = np.full(N+1, d/u) arr[0] = S*np.power(u, N) arr[1:] = arr[0] … Read more
Always use numpy for matrix operations. I’ll assume an m*n numpy array arr. I first did a transpose using the np.transpose function and then I flipped it using the np.fliplr function. output = np.fliplr(np.transpose(arr)) As mentioned in the comments, there is no way to do an in-place replace without a temporary variable for rectangular matrices. … Read more
So the question is how to reorder your columns in a custom way. For example you have the following DF and you want to reorder your columns in the following way (indices): 5, 3, rest… DF In [82]: df Out[82]: A B C D E F G 0 1 0 0.5 5 1 7 6 … Read more
Solution: val_A = [] val_B = [] val_A.append(A1) val_B.append(B1) for i in range(5): val_A.append(val_B[-1] * Z2) val_B.append(val_A[-1] * Z1) Demo: (On a simple test case, to show that it works) # Manual Z1 = 2 Z2 = 3 A1 = 5 B1 = 7 A2 = B1*Z2 B2 = A2*Z1 A3 = B2*Z2 B3 = … Read more
self.tracks seems to be an (n,2) list. The outer loop takes each of these values, turns them two by two into integers x and y, then performs a function/method cv2.circle with an object mask and several other parameters. container[-1] indicates that you want the value of the last index of container. The values of the … Read more
The memory is being managed by your numpy arrays. As soon as they go out of scope (most likely at the end of the PySparse constructor) the arrays cease to exist, and all your pointers are invalid. This applies to both large and small arrays, but presumably you just get lucky with small arrays. You … Read more
ndindex is a convenient way of generating the indices corresponding to a shape: In [33]: arr = np.arange(36).reshape(4,3,3) In [34]: for xy in np.ndindex((3,3)): …: print(xy, arr[:,xy[0],xy[1]]) …: (0, 0) [ 0 9 18 27] (0, 1) [ 1 10 19 28] (0, 2) [ 2 11 20 29] (1, 0) [ 3 12 21 … Read more
To do np.dot last dimension of first matrix must be the same as first dimension of second one. They are not, so you are getting ValueError, that shapes are not aligned. Everything seems to be fine as you printed, but then you forgot about lines: j_temp = np.copy(J_2_SYM) # Add row/col into J_2_SYM j_temp = … Read more
Let’s try this df[“val1”] = df[“val1”].transform(lambda x: x.fillna(x.mean(), limit=2)) df[“val2”] = df[“val2”].transform(lambda x: x.fillna(x.mean(), limit=2)) print df Don’t forget to let us know if it solved your problem 🙂 1 solved how to fill NA with mean only for 2 or less consequective values of NA
Some parts of your question are unclear. It might help to give the context of what you’re trying to achieve, rather than what are the specific steps you’re taking. 1) + 3) In a Normal distribution – fitting the distribution, and estimating the mean and standard deviation – are basically the same thing. The mean … Read more
np.arrays are much faster when using vectorized functions or to say it another way, specifically when you are not iterating over the elements. Here is an example. We will add 1 to each number in the ordered list you created. The setup is similar except I use lists instead of ranges so I can operate … Read more
You have a bunch of 1000 element arrays: In [8]: p.shape Out[8]: (1000,) In [9]: K.shape Out[9]: (1000,) In [10]: R.shape Out[10]: (1000,) In [11]: np.minimum.reduce([p, K, R]).shape Out[11]: (1000,) In [12]: Vl(p).shape Out[12]: (1000,) In [8]: p.shape Out[8]: (1000,) In [9]: K.shape Out[9]: (1000,) In [10]: R.shape Out[10]: (1000,) In [11]: np.minimum.reduce([p, K, R]).shape … Read more
As rafaelc suggested, you can use np.sqrt on a specific column. a = np.array([[1.,2.,3.],[1.,2.,3.]]) a[:,1] = np.sqrt(a[:,1]) a has two rows and three columns. a[:,1] is the second column. 1 solved Using Numpy to update one column by the square root of those column values [closed]