search for: antilog

Dear R-users, I have a basic question about how to determine the antilog of a variable. Say I have some number, x, which is a factor of 2 such that x = 2^y. I want to figure out what y is, i.e. I am looking for the antilog base 2 of x. I have found log2 in the Reference Manual. But I am struggling how to get the antilog of that. Any help will be appreciated! >...

Hi, I am using R for microarray data anlaysis. When I normalize my data, it converts all my data in to log base 2 values. how can I convert back to log base 10..is there any function in R which I can use or how can I take anti log. or is there any function in R for antilog. Please let me know,..if anyone knows.. Thank you so much, Saurin ===== Saurin's WebWorld: http://hawkmail.monmouth.edu/~s0451884

...ummary output from a neg bin GLM? I have 3 significant variables and I can see whether they have a positive or negative effect, but I can't work out how to calculate the magnitude of the effect on the mean of the dependent variable. I used a log link function so I think I might have to use the antilogs of the coefficients but I have no idea how this relates to the dependent variable?? Any help would be much appreciated. My model and output is shown below. Thanks Anna Call: glm.nb(formula = Pass ~ Dist + Time + Wind, data = bats, link = "log", init.theta = 0.8510838809) Dev...

Hi, This might be me missing something painfully obvious but why does the cube root of the following produce an NaN? > (-4)^(1/3) [1] NaN > As we can see: > (-1.587401)^3 [1] -4 Thanks! Greg

...39;t seen anyone post a reply yet so thought I'd throw in my thoughts. I'm no R expert! When you talk about an exponential trend line are you refering to: 1) y=ax^b or 2) y=ae^(bx) If 1) then take base10 logs of y and x and then fit them with simple linear regression. Then calculate the antilog of the residulas and plot these as your trendline. If 2) then take natural logs of y and x and follow the rest of the procedure described in 1). Hope this helps. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richstat at earthlink.net ----- Original Mess...

...Residual 44 1136. 25.83 Total 47 2434. 51.80 * MESSAGE: ratios are based on dispersion parameter with value 1 Dispersion parameter is fixed at 1.00 *** Estimates of parameters *** antilog of estimate s.e. t(*) t pr. estimate Constant -1.1992 0.0211 -56.90 <.001 0.3014 age (23-26) -0.4302 0.0326 -13.20 <.001 0.6504 age (27-65) -0.7999 0.0224 -35.75 <.001 0.4494...

Hi all, I have a dataset which consists of 2 columns. I'd like to plot them on a x-y scatter plot and fit an exponential trendline. I'd like R to determine the equation for the trendline and display it on the graph. Since I am new to R (and statistics), any advice on how to achieve this will be greatly appreciated. Many thanks, Chris -- View this message in context:

...t with x sorted along y. o plot.formula now allows ylab to be set o plotmath had trouble with paste()'ing expressions o plotting math expressions now also works for objects of mode "call" (in particular on the result of substitute()) o locator() on log axes: value antilogged twice o (a real oldie - and trival too) is.recursive now TRUE for expressions o sanitised delete.response o x[["a",]] could crash R o strwidth() crashed R if no device open o predict inconsistencies with offsets straightened out o tapply goofed when FUN...

...t with x sorted along y. o plot.formula now allows ylab to be set o plotmath had trouble with paste()'ing expressions o plotting math expressions now also works for objects of mode "call" (in particular on the result of substitute()) o locator() on log axes: value antilogged twice o (a real oldie - and trival too) is.recursive now TRUE for expressions o sanitised delete.response o x[["a",]] could crash R o strwidth() crashed R if no device open o predict inconsistencies with offsets straightened out o tapply goofed when FUN...

Hi all, Im a little bit confused concerning the effect() command, effects package. I have done several glm models with family=quasipoisson: model <-glm(Y~X+Q+Z,family=quasipoisson) and then used results.effects <-effect("X",model,se=TRUE) to get the "adjusted means". I am aware about the debate concerning adjusted means, but you guys just have to trust me - it