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Top Forums Shell Programming and Scripting How to combing output of cut commands with a delimiter? Post 302901657 by Don Cragun on Wednesday 14th of May 2014 06:34:15 PM
Old 05-14-2014
First, it is not nice to go back and edit your original problem statement after your problem has been solved. Second, your updated problem statement is extremely vague.

What output are you hoping to get from your sample real file? Do you want commas inserted between the fields identified by Col=start_col-end_col? Do you want thirteen lines of output for every input line?
 

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SLASD1(l)								 )								 SLASD1(l)

NAME
SLASD1 - compute the SVD of an upper bidiagonal N-by-M matrix B, SYNOPSIS
SUBROUTINE SLASD1( NL, NR, SQRE, D, ALPHA, BETA, U, LDU, VT, LDVT, IDXQ, IWORK, WORK, INFO ) INTEGER INFO, LDU, LDVT, NL, NR, SQRE REAL ALPHA, BETA INTEGER IDXQ( * ), IWORK( * ) REAL D( * ), U( LDU, * ), VT( LDVT, * ), WORK( * ) PURPOSE
SLASD1 computes the SVD of an upper bidiagonal N-by-M matrix B, where N = NL + NR + 1 and M = N + SQRE. SLASD1 is called from SLASD0. A related subroutine SLASD7 handles the case in which the singular values (and the singular vectors in factored form) are desired. SLASD1 computes the SVD as follows: ( D1(in) 0 0 0 ) B = U(in) * ( Z1' a Z2' b ) * VT(in) ( 0 0 D2(in) 0 ) = U(out) * ( D(out) 0) * VT(out) where Z' = (Z1' a Z2' b) = u' VT', and u is a vector of dimension M with ALPHA and BETA in the NL+1 and NL+2 th entries and zeros else- where; and the entry b is empty if SQRE = 0. The left singular vectors of the original matrix are stored in U, and the transpose of the right singular vectors are stored in VT, and the singular values are in D. The algorithm consists of three stages: The first stage consists of deflating the size of the problem when there are multiple singular values or when there are zeros in the Z vector. For each such occurence the dimension of the secular equation problem is reduced by one. This stage is performed by the routine SLASD2. The second stage consists of calculating the updated singular values. This is done by finding the square roots of the roots of the secular equation via the routine SLASD4 (as called by SLASD3). This routine also calculates the singular vectors of the current problem. The final stage consists of computing the updated singular vectors directly using the updated singular values. The singular vectors for the current problem are multiplied with the singular vectors from the overall problem. ARGUMENTS
NL (input) INTEGER The row dimension of the upper block. NL >= 1. NR (input) INTEGER The row dimension of the lower block. NR >= 1. SQRE (input) INTEGER = 0: the lower block is an NR-by-NR square matrix. = 1: the lower block is an NR-by-(NR+1) rectangular matrix. The bidiagonal matrix has row dimension N = NL + NR + 1, and column dimension M = N + SQRE. D (input/output) REAL array, dimension (N = NL+NR+1). On entry D(1:NL,1:NL) contains the singular values of the upper block; and D(NL+2:N) contains the singular values of the lower block. On exit D(1:N) contains the singular values of the modified matrix. ALPHA (input) REAL Contains the diagonal element associated with the added row. BETA (input) REAL Contains the off-diagonal element associated with the added row. U (input/output) REAL array, dimension(LDU,N) On entry U(1:NL, 1:NL) contains the left singular vectors of the upper block; U(NL+2:N, NL+2:N) contains the left singular vectors of the lower block. On exit U contains the left singular vec- tors of the bidiagonal matrix. LDU (input) INTEGER The leading dimension of the array U. LDU >= max( 1, N ). VT (input/output) REAL array, dimension(LDVT,M) where M = N + SQRE. On entry VT(1:NL+1, 1:NL+1)' contains the right singular vectors of the upper block; VT(NL+2:M, NL+2:M)' contains the right singular vectors of the lower block. On exit VT' contains the right singular vectors of the bidiagonal matrix. LDVT (input) INTEGER The leading dimension of the array VT. LDVT >= max( 1, M ). IDXQ (output) INTEGER array, dimension(N) This contains the permutation which will reintegrate the subproblem just solved back into sorted order, i.e. D( IDXQ( I = 1, N ) ) will be in ascending order. IWORK (workspace) INTEGER array, dimension( 4 * N ) WORK (workspace) REAL array, dimension( 3*M**2 + 2*M ) INFO (output) INTEGER = 0: successful exit. < 0: if INFO = -i, the i-th argument had an illegal value. > 0: if INFO = 1, an singular value did not converge FURTHER DETAILS
Based on contributions by Ming Gu and Huan Ren, Computer Science Division, University of California at Berkeley, USA LAPACK version 3.0 15 June 2000 SLASD1(l)
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