a = np.arange(18).reshape(9,2) b = a.reshape(3,3,2).swapaxes(0,2) # a: array([[ 0, 1], [ 2, 3], [ 4, 5], [ 6, 7], [ 8, 9], [10, 11], [12, 13], [14, 15], [16, 17]]) # b: array([[[ 0, 6, 12], [ 2, 8, 14], [ 4, 10, 16]], [[ 1, 7, 13], [ 3, 9, 15], [ 5, … Read more
Yes, it does behave like -1 in numpy.reshape(), i.e. the actual value for this dimension will be inferred so that the number of elements in the view matches the original number of elements. For instance: import torch x = torch.arange(6) print(x.view(3, -1)) # inferred size will be 2 as 6 / 3 = 2 # … Read more
First I convert lower values of df to NaN by where and numpy.triu and then stack, reset_index and set column names: import numpy as np print df a b c a 1.0 0.5 0.3 b 0.5 1.0 0.4 c 0.3 0.4 1.0 print np.triu(np.ones(df.shape)).astype(np.bool) [[ True True True] [False True True] [False False True]] df … Read more
Use torch.unsqueeze(input, dim, out=None): >>> import torch >>> a = torch.Tensor([1, 2, 3, 4, 5]) >>> a 1 2 3 4 5 [torch.FloatTensor of size 5] >>> a = a.unsqueeze(0) >>> a 1 2 3 4 5 [torch.FloatTensor of size 1×5]
Wrap the condition in which: df[which(df$number1 < df$number2), ] How it works: It returns the row numbers where the condition matches (where the condition is TRUE) and subsets the data frame on those rows accordingly. Say that: which(df$number1 < df$number2) returns row numbers 1, 2, 3, 4 and 5. As such, writing: df[which(df$number1 < df$number2), … Read more