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  • matrix[0][i] = 1.; for (int i = 0; i < markers - 1; i++)
    654 bytes (81 words) - 09:24, 19 January 2017
  • // i ranges from 0 .. individuals - 1 // j ranges from 0 .. markers - 1
    2 KB (162 words) - 15:56, 13 September 2013
  • RetrieveMemoryBlock(markers - 1); SampleHaplotypes(leftMatrices[markers - 1], first, second, rand);
    1 KB (131 words) - 14:11, 2 October 2013
  • right[0][0] = (1. - theta) * (1. - theta); right[1][0] = right[0][1] = (1. - theta) * theta / states;
    1 KB (110 words) - 15:55, 2 October 2013
  • ...gree files, MaCH can accept pedigree files that encode bases A, C, G, T as 1, 2, 3, 4. Here is an example: FAM1001 ID1234 0 0 M 1 1 1 2 2 2
    649 bytes (96 words) - 10:24, 4 June 2010
  • RetrieveMemoryBlock(markers - 1); ImputeGenotypes(leftMatrices[markers - 1], markers - 1);
    722 bytes (69 words) - 14:19, 2 October 2013
  • 1994 5 1 2 1 4.05563 0.839037 2.73226 345 1 0 0 1 3.45514 0.785963 1.95526
    2 KB (193 words) - 05:10, 17 February 2015
  • while (cache[h] != -1 && cache[h] != key) h = h + 1 >= cacheSize ? 0 : h + 1;
    268 bytes (29 words) - 13:58, 28 October 2013
  • return (first * (first + 1) / 2 + second); return (second * (second + 1) / 2 + first);
    513 bytes (50 words) - 16:03, 2 October 2013
  • Ref ID: 0 Ref Name: 1 NumRecs: 4 Ref ID: -1 indicates unmapped (Ref Name will be *)
    2 KB (271 words) - 17:54, 6 January 2014
  • ...cles for this position. There are numBases cycles, separated by a ':'. (-1 represents the cycle of a deletion) ...sition. Sequence of numBases 0's and 1's. 0 represents forward strand and 1 represents reverse strand.
    5 KB (646 words) - 17:39, 3 January 2014
  • return -1; if (mask[i] == '1')
    500 bytes (64 words) - 13:39, 28 October 2013
  • {|border="1" cellspacing="0" cellpadding="2" ...Depth at Site || INFO:DP < #samples * FILTER_MIN_SAMPLE_DP > 0 || #samples*1
    4 KB (641 words) - 10:53, 29 October 2014
  • X5 X5 0 -9 0 0 0 2 2 1 2 2 2 1 10 1:10 A T
    3 KB (398 words) - 17:01, 19 February 2013
  • lastOccurrence = sortedPositions.Length() - 1; firstMax = probe - 1;
    1 KB (90 words) - 16:09, 28 October 2013
  • {| class="wikitable" style="width:100%" border="1" | Line is at least 2 characters long ('@' and at least 1 for the sequence identifier)
    3 KB (529 words) - 14:01, 7 September 2011
  • ...ue implies an excess of homozygotes. <math>F_{IC}</math> ranges from -1 to 1. F_{IC} & = 1 - \frac{O[Het]}{E[Het|\textbf{p}]} \\
    2 KB (404 words) - 15:53, 3 August 2015
  • * Phase 1 LungGO - ccds_2008 * Phase 1 WHISPGO - BMI/T2D - ccds_2008
    918 bytes (134 words) - 21:44, 14 September 2010
  • ...art of the 2nd read. (mate0BasedClippedStart - 0BasedPositionClippedEnd - 1) * chromosome is unknown (-1/*)
    5 KB (728 words) - 01:45, 6 March 2016
  • double ptotal = P[1] + P[2] + P[3] + P[4]; int mle = 1;
    1 KB (165 words) - 14:50, 25 September 2013

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