datS) has a consistently high ranking (or low ranking) according to multiple ordinal scores
(corresponding to the columns of the input data frame datS).rankPvalue(datS, columnweights = NULL,
na.last = "keep", ties.method = "average",
calculateQvalue = TRUE, pValueMethod = "all")datS represents an
ordinal variable (which can take on negative values). The columns correspond to (possibly signed) object
significance measures, e.g., statistics datZ.
If it is set to NULL then all weights are equal.TRUE, missing values in the data are
put last (i.e. they get the highest rank values). If FALSE, they are put first;
if NA, they are removed; irank for
more details.datoutrank and
datoutscale,:Example: Imagine you have 5 vectors of Z statistics corresponding to the columns of datS. Further assume that a gene has ranks 1,1,1,1,20 in the 5 lists. It seems very significant that the gene ranks number 1 in 4 out of the 5 lists. The function rankPvalue can be used to calculate a p-value for this occurrence.
The function uses the central limit theorem to calculate asymptotic p-values for two types of test statistics that measure consistently high or low ordinal values. The first method (referred to as percentile rank method) leads to accurate estimates of p-values if datS has at least 4 columns but it can be overly conservative. The percentile rank method replaces each column datS by the ranked version rank(datS[,i]) (referred to ask low ranking) and by rank(-datS[,i]) (referred to as high ranking). Low ranking and high ranking allow one to find consistently small values or consistently large values of datS, respectively. All ranks are divided by the maximum rank so that the result lies in the unit interval [0,1]. In the following, we refer to rank/max(rank) as percentile rank. For a given object (corresponding to a row of datS) the observed percentile rank follows approximately a uniform distribution under the null hypothesis. The test statistic is defined as the sum of the percentile ranks (across the columns of datS). Under the null hypothesis that there is no relationship between the rankings of the columns of datS, this (row sum) test statistic follows a distribution that is given by the convolution of random uniform distributions. Under the null hypothesis, the individual percentile ranks are independent and one can invoke the central limit theorem to argue that the row sum test statistic follows asymptotically a normal distribution. It is well-known that the speed of convergence to the normal distribution is extremely fast in case of identically distributed uniform distributions. Even when datS has only 4 columns, the difference between the normal approximation and the exact distribution is negligible in practice (Killmann et al 2001). In summary, we use the central limit theorem to argue that the sum of the percentile ranks follows a normal distribution whose mean and variance can be calculated using the fact that the mean value of a uniform random variable (on the unit interval) equals 0.5 and its variance equals 1/12.
The second method for calculating p-values is referred to as scale method. It is often more powerful but its asymptotic p-value can only be trusted if either datS has a lot of columns or if the ordinal scores (columns of datS) follow an approximate normal distribution. The scale method scales (or standardizes) each ordinal variable (column of datS) so that it has mean 0 and variance 1. Under the null hypothesis of independence, the row sum follows approximately a normal distribution if the assumptions of the central limit theorem are met. In practice, we find that the second approach is often more powerful but it makes more distributional assumptions (if datS has few columns).
Storey JD and Tibshirani R. (2003) Statistical significance for genome-wide experiments. Proceedings of the National Academy of Sciences, 100: 9440-9445.
rank, qvalue