search for: p_value

Dear All: I create a file named :"P_Value" with only one simple function: P_Value <- function( Table ) { S = fisher.test(Table, alternative = "two.sided"); return(S$p.value); } However, it seems that it's impossible to use this function directly, because no matter...

...ect, making this an entirely within-subject design. I would like an output that compares the significance of the 3 levels of difficulty for each condition, as well as the overall interaction of condition*difficulty. The ideal output would look like this: condition1:diff25 vs. condition1:diff50 p_value = .... condition1:diff25 vs. condition1:diff75 p_value = .... condition1:diff50 vs. condition1:diff75 p_value = .... condition2:diff25 vs. condition1:diff50 p_value = .... condition2:diff25 vs. condition1:diff75 p_value = .... condition2:diff50 vs. condition1:diff75 p_value = .... condition3...

Hi, I am using an if else statement inside a function ?. If I use that function I have no problems ?. If I use the function with the if else statement inside a second function I get the following waring: Warning message: In if (pval == 0) p_value <- "< 2.2e-16" else p_value <- pval : the condition has length> 1 and only the first element will be used Using the second function I get the expected results, with a real p-value even if it is extremely small, or " _____________________________________________________...

Hello Friends, I am new to R and stuck with a problem. i have two columns drug_A and drug_B, i have plotted a scatter plot using the ggplot2 function. My problem is with the third column, it is the p-value column. I want to color and size points differently based on the p_value, the p_value range is between 0<0.0001< 0.001<0.05<1. I used a script using the ifelse loop, but it considers only two conditions at one time. Please tell me a way to fulfill all the conditions. here is the script: library(lattice) qplot(Drug_A, Drug_B, data=dsmall, size = p_value,...

...o define a function that will crop the p-values from an Anova (so that I can use the function with a 'for loop' later on to go through all the genes): > p.fun <- function(arg) { two_way_anova <- aov(arg ~ age * treatment, data = example.df) two_way_sum <- summary(two_way_anova) p_values <- two_way_sum[[1]]$"Pr(>F)"[1:3] return(p_values) } Unfortunately my setup seems to be flawed as I'm not capable to call my function: > p.fun(gene1) Error in eval(expr, envir, enclos) : object 'gene1' not found > p.fun("gene1") Error in model.frame.def...

Buena tarde, Estoy llevando a cabo un trabajo y no encuentro la forma de llenar una matriz con el p_value y un estadístico calculado. Un poco mas detallado, tengo muestras aleatorias, calculo por ejemplo la prueba t, wilcoxon, etc y requiero llevar por una parte los p_value de k muestras a una matriz creada anteriormente y por otro lado el valor del estadístico de las mismas k muestras a otra matriz cr...

...des hacer el cálculo de una forma mucho más compacta y rápida. Esta forma es especialmente recomendable cuando tienes muchas columnas y muchas filas. > library(data.table) > myDT <- as.data.table(mtcars) > myDTlong <- melt(myDT, measure.vars=1:ncol(myDT)) > myDTlong[ , list(p_value = shapiro.test(value)$p.value, v_stat = shapiro.test(value)$statistic) , by = .(variable)] variable p_value v_stat 1: mpg 1.228814e-01 0.9475647 2: cyl 6.058338e-06 0.7533100 3: disp 2.080657e-02 0.9200127 4: hp 4.880824e-02 0.9334193 5: drat 1.100608e-01 0....

Hi. how can i insert the value of p_value from KS.test into vector? Regards, Irit. -- View this message in context: http://r.789695.n4.nabble.com/retrieveing-value-from-KS-test-tp4663439.html Sent from the R help mailing list archive at Nabble.com.

I have csv files imported in r each with 2 columns and many many rows. I have sorted the data in them but want to extract some values. The first column is an ID The second is a p-value ( now sorted in increasing order with NA's last) I want to extract the rows with a p-value of less than 0.05) What commands would help the table is called AnovaSort with column headings MCI & p-value Many

This was working a few weeks ago, but perhaps the package has been updated since then. model.1 <- lrm(response ~ p_value, data=c_abl_oncogene_1_RTK) When I run the following command . . . . prediction.1 <- predict(model.1, type=c("fitted")) I get the following error message. . . . Error in predictDesign(object, ..., type = "lp", se.fit = FALSE) : could not find function "Varcov" I...

...ble.cnts)), lambda) variable.cnts       <-     c(variable.cnts, 0)   variable.cnts.prs <-     c(variable.cnts.prs, 1-sum(variable.cnts.prs)) tst                    <-     chisq.test(variable.cnts, p=variable.cnts.prs)   chi_squared        <-     as.numeric(unclass(tst)$statistic) p_value             <-     as.numeric(unclass(tst)$p.value) df                     <-     tst[2]$parameter     cv1                    <-     qchisq(p=.01, df=tst[2]$parameter, lower.tail = FALSE, log.p = FALSE)   cv2                    <-     qchisq(p=.05, df=tst[2]$parameter, lower.tail...

...ble.cnts)), lambda) variable.cnts ????? <-???? c(variable.cnts, 0) ? variable.cnts.prs <-???? c(variable.cnts.prs, 1-sum(variable.cnts.prs)) tst ?????????????????? <-???? chisq.test(variable.cnts, p=variable.cnts.prs) ? chi_squared ?????? <-???? as.numeric(unclass(tst)$statistic) p_value? ?????????? <-???? as.numeric(unclass(tst)$p.value) df ??????????????????? <-???? tst[2]$parameter ? ? cv1??????????????????? <-???? qchisq(p=.01, df=tst[2]$parameter, lower.tail = FALSE, log.p = FALSE) ? cv2??????????????????? <-???? qchisq(p=.05, df=tst[2]$parameter, lower.tail...

...ble.cnts)), lambda) variable.cnts       <-     c(variable.cnts, 0)   variable.cnts.prs <-     c(variable.cnts.prs, 1-sum(variable.cnts.prs)) tst                    <-     chisq.test(variable.cnts, p=variable.cnts.prs)   chi_squared        <-     as.numeric(unclass(tst)$statistic) p_value             <-     as.numeric(unclass(tst)$p.value) df                     <-     tst[2]$parameter     cv1                    <-     qchisq(p=.01, df=tst[2]$parameter, lower.tail = FALSE, log.p = FALSE)   cv2                    <-     qchisq(p=.05, df=tst[2]$parameter, lower.tail...

...), lambda) variable.cnts ????? <-???? c(variable.cnts, 0) ? variable.cnts.prs <-???? c(variable.cnts.prs, 1-sum(variable.cnts.prs)) tst ?????????????????? <-???? chisq.test(variable.cnts, p=variable.cnts.prs) ? chi_squared ?????? <-???? as.numeric(unclass(tst)$statistic) p_value? ?????????? <-???? as.numeric(unclass(tst)$p.value) df ??????????????????? <-???? tst[2]$parameter ? ? cv1??????????????????? <-???? qchisq(p=.01, df=tst[2]$parameter, lower.tail = FALSE, log.p = FALSE) ? cv2??????????????????? <-???? qchisq(p=.05, df=tst[2]$parameter, lo...

Buen día estimados Estoy tratando de hacer un tibble con los resultados de un apply que se muestran en la consola que me da R, no estoy seguro si eso se pueda hacer, pero me gustaría organizar los resultados de esa manera. mi código es: data("mtcars") Mtcars_matriz <- as.matrix(mtcars) apply(Mtcars_matriz, MARGIN =2, FUN = shapiro.test) DF2 <- tibble(Variable = NA, W = NA, Pvalue =

...e approach you proposed. As you suggested, imagine that the model we end up after the model selection procedure is: mod2.1 <- lmer(dT_purs ~ T + Z + (1 +T+Z| subject), data =x, REML= FALSE) According to the common procedures specified in many manuals and recent papers, if I want to compute the p_values relative to each term, I will perform a likelihood test, in which the deviance of the (-2LL) of a model containing the specific term is compared to another model without it. In the case of the fixed effect terms I have no problem in the interpretation of the results. Each comparison returns a sign...

...nction in the pheno package. It seems to be set automatically at 0.05 and there seems to be neither an input function to set a different level, as there are with some changepoint functions, nor an output to give one the p-value of the changepoint identification which you get with functions ie func$p_value etc ... Thanks Nick Wray [[alternative HTML version deleted]]

...nG3 * ((G3Mean - GrandMean)^2))/2 } An <- anova(lm(Var ~ Group)) MSwithin = An[3]['Residuals',] DegF = An$Df[length(An$Df)] Fval = MScontr / MSwithin Pval = 1 - pf(Fval, 1, DegF) return (list(MS_contrasts = MScontr, MS_within = MSwithin, F_value = Fval, P_value = Pval)) } ## The first two contrasts produce the same (+/- rounding error) ## p-values as obtained using contrasts() MyContrast(sugars$length, sugars$treatment, 'control', c("fructose", "gluc+fruct", "glucose", "sucrose")) MyContrast(suga...

Hi All, I was wondering if someone could help me to solve this issue with lmer. In order to understand the best mixed effects model to fit my data, I compared the following options according to the procedures specified in many papers (i.e. Baayen <http://www.google.it/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CDsQFjAA

Hi, I have been examining large data and need to do simple linear regression with the data which is grouped based on the values of a particular attribute. For instance, consider three columns : ID, x, y, and I need to regress x on y for each distinct value of ID. Specifically, for the set of data corresponding to each of the 4 values of ID (76,111,121,168) in the below data, I should invoke