transpose

I have data that looks like a b c 1 5 4 3 6 1 2 5 3 I want to convert it to convert all the columns to rows and want an output like r1 r2 r3 r4 a 1 3 2 b 5 6 5 c 4 1 3 Thanks in advance We can transpose the dataset and convert to data.frame with the first column as the row names. m1 <- t(df1) d2 <- data.frame(r1= row.names(m1), m1, row.names=NULL) EDIT: Included the row.names argument in the data.frame call (from @Richard Scriven's comment) Or as @Ananda Mahto mentioned, we can use names(df1) to create the 'r1' column, thereby skipping the creation of an object in the global environment. d2 <-

if A matrix has complex element and I want to transpose A to A' using the command >>A' Why it is design that a+bi be transformed into a-bi ? What it use for? From here : for complex matrices, it is almost always the case that the combined operation of taking the transpose and complex conjugate arises in physical or computation contexts and virtually never the transpose in isolation ( Strang 1988 , pp. 220-221). In matlab if you want to transpose without conjugating use .' . Actually I'd argue that there are deep reasons why the transpose IS the conjugate . Consider a matrix representation of

This question already has an answer here: Transpose list of lists 10 answers I have a 2d list like this: 1 2 3 4 5 6 and I want to make this: 1 4 2 5 3 6 I've tried to do a for loop and switch each value but I keep getting an index out of bound error. Here's what I have: for i in results: for j in range(numCenturies): rotated[i][j] = results [j][i] aga From python documentation on zip function: This function returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables. The returned list is truncated in length to the length of the

I have a matrix I am trying to transpose in R but the t() function does not return the right answer. How can I transpose the matrix? > xx=matrix(c(3,7,4,8),2,byrow=TRUE) > xx [,1] [,2] [1,] 3 7 [2,] 4 8 > t(xx) [1] 0.7071068 0.7071068 Josh O'Brien This answer is incorrect, but in ways that were enlightening to me and might be to others, so I'll leave it up. As @mnel noted, the base R function t() must be masked by another function of the same name. Try removing the function t() and doing t(xx) again. I guarantee you'll get the correct results. What do you get when you run this: rm(t) t(xx) If

I have a 3D matrix x_test of size (100, 33, 66) and I want to change its dimensions to (100, 66, 33) . What is the most efficient way to do this using python3.5? I look for something along those lines: y = x_test.transpose() You can pass the desired dimensions to the function np.transpose using in your case np.transpose(x_test, (0, 2, 1)) . For example, import numpy as np x_test = np.arange(30).reshape(3, 2, 5) print(x_test) print(x_test.shape) This will print [[[ 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] [25 26 27 28 29]]] (3, 2, 5) Now, you can transpose

Intel has included __MM_TRANPOSE4_PS to transpose a 4x4 matrix of vectors. I'm wanting to do the equivalent with __m256d. However, I can't seem to figure out how to get _mm256_shuffle_pd in the same manner. _MM_TRANSPOSE4_PS Code #define _MM_TRANSPOSE4_PS(row0, row1, row2, row3) { \ __m128 tmp3, tmp2, tmp1, tmp0; \ \ tmp0 = _mm_shuffle_ps((row0), (row1), 0x44); \ tmp2 = _mm_shuffle_ps((row0), (row1), 0xEE); \ tmp1 = _mm_shuffle_ps((row2), (row3), 0x44); \ tmp3 = _mm_shuffle_ps((row2), (row3), 0xEE); \ \ (row0) = _mm_shuffle_ps(tmp0, tmp1, 0x88); \ (row1) = _mm_shuffle_ps(tmp0, tmp1, 0xDD); \