# [R] field significance test

Jim Lemon drj|m|emon @end|ng |rom gm@||@com
Mon Sep 6 10:41:43 CEST 2021

```HI Ani,
I would create these two matrices:

# matrix of logicals for positive stat values
posvalue<-df3 > 0
# matrix of logicals for significance
sigstat<-df4 < 0.05

Then you can identify the positive/negative and significant values:

which(posvalue & sigstat)
[1] 12
which(!posvalue & sigstat)
[1] 20 76 78

and as you note, column 2 has 2 significant results, one statistical
value positive and the other negative.
I'm not sure what sort of histogram you want, perhaps all ten columns
with groups of ten bars for each column (very messy and sparse). Maybe
a bit more info will enlighten me.

Jim

On Mon, Sep 6, 2021 at 4:37 PM ani jaya <gaaauul using gmail.com> wrote:
>
> Dear r-list member,
>
> I want to plot a histogram that shows a number of station that have a
> significant statistic (positive or negative) based on the value itself
> and its p-value. df3 shows the test statistic value (column shows the
> station and rows show the result from the resample matrix
> (repetition/bootstrap)) and df4 shows the p-value.
>
> #the value
> structure(c(0.569535339474781, 1.02925697755861, 1.08125714350978,
> 0.50589479161552, -0.695827095264809, 0.455608022735733, 1.2552019505074,
> 0.981335144120386, 1.63020923423253, -0.424613279862939, 0.429207234903993,
> 1.99059339634301, -1.25731480224036, 0.64293796635093, 0.0189774621961392,
> 0.1163965630274, -1.41756397958877, 1.58945674395921, -1.2551489541395,
> -2.84122761058959, -0.72446669544026, -0.719331298629362, -0.164045813998067,
> 0.444120153507258, -0.0845757313567553, -0.27732982718919, -0.166982066770785,
> -0.193859909749249, 0.277426534878283, -0.0430460496295642, -0.0741475736028902,
> -0.017026178205196, 0.732589091697401, 0.332813962514037, -0.0860983232517636,
> 0.155930932436498, -0.438635444604027, 0.046881008364722, -0.704876076807635,
> -0.945506782070735, 0.662399207637722, -0.860903464600488, 1.06638547921749,
> -0.462184163508299, 0.442447468362937, 0.145655792120232, 0.696309974316211,
> 1.84692085953474, 0.00841868461519582, -1.04408256815264, -0.548599461573869,
> 1.22352273108675, 0.0191993545723452, 1.26090162037733, 0.192106046362172,
> -1.02864978106213, -0.0712068006002629, -0.674610175422543, -0.658383381010154,
> -1.52779151484935, 0.479809528798632, -0.112078644619679, -0.19482661081522,
> -0.192179943664117, -0.246553759113406, -0.563554156777087, -1.0236492805268,
> 0.0289772842372375, -0.274878506644853, 0.95578159001869, -0.27550722692588,
> -0.66586322268903, 1.24703690613745, -0.00368775734780707, -0.0766884108214613,
> -1.41610325144406, 0.518897523428314, -2.12289477996499, 0.968369305561191,
> 0.0766656793804207, 0.470712743077857, 0.241711948576043, 0.0636131491007723,
> -1.13735866614159, 0.625015831730259, -0.234696421716696, 0.358555918256736,
> -0.651761882852838, -0.236796663592383, 0.0421395303375618, 0.574747610964774,
> -0.730646230622174, -0.20839489662388, -1.4832025994155, -0.366841536561336,
> 0.621868015281511, 0.945609952617796, 0.297055307072896, 0.737974050847397,
> 1.49862070675738), .Dim = c(10L, 10L))
>
> #the p-value
> structure(c(0.560903574193679, 0.358019718822816, 0.320136568444488,
> 0.721538652049639, 0.419898899237915, 0.511481779449553, 0.208829636238898,
> 0.535905791761543, 0.252523383923989, 0.721538652049639, 0.487651926831611,
> 0.0281856103410957, 0.138370395238992, 0.639104270712721, 0.98503410973661,
> 0.955123383216192, 0.358019718822816, 0.138370395238992, 0.252523383923989,
> 0.0373292396736942, 0.302215769747998, 0.302215769747998, 0.807343273858921,
> 0.560903574193679, 0.955123383216192, 0.836526366120417, 0.807343273858921,
> 0.807343273858921, 0.693640621783759, 0.895532903167044, 0.895532903167044,
> 0.98503410973661, 0.159470497055087, 0.560903574193679, 0.925275729900227,
> 0.865936215436343, 0.441845502530452, 0.98503410973661, 0.358019718822816,
> 0.170893484254114, 0.586452625432322, 0.268412562734209, 0.102689728987727,
> 0.511481779449553, 0.666151798537229, 0.925275729900227, 0.358019718822816,
> 0.0581501553999165, 0.98503410973661, 0.170893484254114, 0.586452625432322,
> 0.464434476654839, 0.98503410973661, 0.252523383923989, 0.925275729900227,
> 0.377977518007105, 0.98503410973661, 0.586452625432322,
> 0.666151798537229, 0.284975267823252, 0.560903574193679,
> 0.721538652049639, 0.778425914188847,
> 0.836526366120417, 0.778425914188847, 0.511481779449553, 0.087825095630195,
> 0.98503410973661, 0.693640621783759, 0.208829636238898, 0.807343273858921,
> 0.222740206090239, 0.222740206090239, 0.98503410973661, 0.925275729900227,
> 0.0373292396736942, 0.586452625432322, 0.00322938266821475, 0.222740206090239,
> 0.865936215436343, 0.338738311334395, 0.639104270712721, 0.895532903167044,
> 0.0533495868962313, 0.268412562734209, 0.721538652049639, 0.721538652049639,
> 0.195559652706897, 0.778425914188847, 0.880692897134707, 0.398606385377039,
> 0.398606385377039, 0.693640621783759, 0.102689728987727, 0.666151798537229,
> 0.252523383923989, 0.358019718822816, 0.778425914188847, 0.284975267823252,
> 0.0633043080023749), .Dim = c(10L, 10L))
>
> #find the positive significant station
> df5<-df3
> df5[df4>0.05|df5<0]<-NA
> df5[df5>0]<-1
> pos<-as.numeric(rowSums(df5, na.rm=T))
> hist(pos)
>
> #find the negative significant station
> df6<-df3
> df6[df4>0.05|df5>0]<-NA
> df6[df6<0]<-1
> neg<-as.numeric(rowSums(df6, na.rm=T))
> hist(neg)
>
> but above code is not correct because the 0 station (row when there is
> no significant station detected) should be the same. The problem is
> when the row produces significant positive and negative at the same
> time. Is there any way to combine positive and negative significant
> value and plot the histogram? or we can calculate the 0 station first
> separately?
>
> Any lead is really appreciated. Thank you.
>
> Ani Jaya
>
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