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May 17

Suggestions for extracting data from a damaged Excel file?

Hello all. I have an Excel file that I put a lot of work into and it's damaged. I saved it recently and when I went back to work on it, it's name had changed to have this in front of the original file name: ~$. when I try to open it, Excel says: "Excel cannot open this file. The file might have been damaged or modified from its original format". I was just wondering if there's anything anyone might suggest I try to extract the data from the file. I am very far from a computer wiz. It's not the end of the world, but losing it is a lot of work down the drain. At the same time, it contains sensitive information, and I would not want to send it to an expert to open it. So if anyone has some spoonfed instructions... I am on an IMac running High Sierra. Thanks!--185.230.124.52 (talk) 03:36, 17 May 2018 (UTC)[reply]

Someone else probably has a better answer, but what format is it in? XLS? Bubba73 ^{You talkin' to me?} 04:19, 17 May 2018 (UTC)[reply]

It's xlsx.--67.244.114.239 (talk) 04:30, 17 May 2018 (UTC)[reply]

The ~$ at the front means that this is a temporary file resulting from an improper closedown of Excel (or the operating system). I assume that you've checked that there is not also the original filename somewhere. (Also check that you are not still running another copy of Excel.) Dbfirs 07:45, 17 May 2018 (UTC)[reply]

PAIRWISE SUMMATION

PAIRWISE SUMMATIONS I do a lot of computations, mostly in complex numbers. That includes multiplications and additions. Rounding errors accumulate and eventually become a problem. I found this article in Wikipedia [1] but it gives a pseudocode instead of C or C++ and it seems some information is missing.

I quote:

In pseudocode, the pairwise summation algorithm for an array x of length n > 0 can be written:

s = pairwise(x[1…n])
      if n ≤ N                    base case: naive summation for a sufficiently small array
          s = x[1]
          for i = 2 to n
              s = s + x[i]
      else                        divide and conquer: recursively sum two halves of the array
          m = floor(n / 2)
          s = pairwise(x[1…m]) + pairwise(x[m+1…n])
      endif

What is clear so far that the algorithm is recursive because the function calls upon itself, and then the input is an array of reals but I mostly operate with complex numbers. So, this is my thought on how it could be implemented in C, my OS is Linux Ubuntu:

dcomp is my definition of double complex. Suppose I have an array of 100 complex numbers to add, my typical situation. I expect to get one complex number in the end.

typedef std::complex<double> dcomp;

int main () {

int n = 100;
 dcomp ss, * xx;         // ss is the expected sum
 xx = new dcomp [n];     // array of input complex numbers, size of the array is n
 ss = pairwise (0, n, xx);

}
public : dcomp pairwise (int start, int end, dcomp * xx)
{

int mm;
 dcomp ss;
 if (n <= N) {
   ss = xx[0];
   for (int ii = 1; ii < n; ii++) 
     ss += xx[ii];
  }
  else {
    mm = n/2;
    ss = pairwise(0,mm,xx) + pairwise(mm+1,n,xx);
  }
  return ss;

} // pairwise

I am not sure this code in this form will work. First, N is undefined. Obviously, N is an integer. Any hunch as to its possible value? My hunch is that iT is something like $N<<n$ , but how is it defined? Also is it a number that changes during iterations? I must make this code work. I do have considerable loss of precision in my computations of significant practical value. I will appreciate any comments to improve the code. Thanks, - AboutFace 22 (talk) 16:16, 17 May 2018 (UTC)[reply]

The pseudocode clearly states that N is small enough to make "a sufficiently small array". The point is that you have to have a point at which you don't make a recursive call. You start adding. How small of an array do you think you need to be sure that you won't have rounding errors? That is how small N needs to be. Example: N=5. Then, when the array has 5 (or less) elements, you will just add them together and not perform a recursive call. 209.149.113.5 (talk) 18:56, 17 May 2018 (UTC)[reply]

Thank you. It makes sense. I will follow it, although it is unclear how it operates. I will think about it too. AboutFace 22 (talk) 22:28, 17 May 2018 (UTC)[reply]

You might also be interested in Kahan summation algorithm. Dmcq (talk) 22:57, 17 May 2018 (UTC)[reply]

I am familiar with it but it's considered much slower than pairwise and only slightly better in avoiding rounding errors. AboutFace 22 (talk) 19:45, 18 May 2018 (UTC)[reply]