r - How to remove outliers from a list of vectors? -


i have list of vectors : 

tdatm.sp=structure(list(x3co = c(24.88993835, 25.02366257, 24.90308762 ), x3cs = c(25.70629883, 25.26747704, 25.1953907), x3cd = c(26.95723343,  26.84725571, 26.2314415), x3csd = c(36.95250702, 36.040905, 36.90475845 ), x5co = c(25.44123077, 24.97585869, 24.86075592), x5cs = c(25.71570396,  26.10244179, 25.39032555), x5cd = c(27.67508507, 27.18985558,  26.93682098), x5csd = c(36.26528549, 34.88553238, 33.97910309 ), x7co = c(24.7142601, 24.08443642, 23.97057915), x7cs = c(24.55734444,  24.56562042, 24.7589817), x7cd = c(27.14260101, 26.65704346,  26.49533081), x7csd = c(33.89881897, 32.91091919, 32.79199219 ), x9co = c(26.86141014, 26.42648888, 25.8350563), x9cs = c(28.17367744,  27.27400589, 26.58813667), x9cd = c(28.88915062, 28.32597542,  28.2713623), x9csd = c(34.61352158, 35.84189987, 35.80329132)), .names = c("x3co",  "x3cs", "x3cd", "x3csd", "x5co", "x5cs", "x5cd", "x5csd", "x7co",  "x7cs", "x7cd", "x7csd", "x9co", "x9cs", "x9cd", "x9csd"))  > head(tdatm.sp) $x3co [1] 24.88994 25.02366 24.90309  $x3cs [1] 25.70630 25.26748 25.19539  $x3cd [1] 26.95723 26.84726 26.23144  $x3csd [1] 36.95251 36.04091 36.90476  $x5co [1] 25.44123 24.97586 24.86076  $x5cs [1] 25.71570 26.10244 25.39033

i remove outliers each individual vector using hampel method.

one way found :

repoutliers=function(x){ med=median(x); mad=mad(x); x[x>med+3*mad | x<med-3*mad]=na; return(x)} lapply(tdatm.sp, repoutliers)

but wondering if possible without declaring new function, directly within lapply. lapply sends each individual vector function repoutliers, know how operate on individual vectors directly within lapply ? let's swap repoutliers function "replace", same word calling individual vectors in arguments of replace (lapply(x,fun,...); ... = replace arguments).

in brief : how manipulate individual vectors lapply sends function winthin lapply ?

it's more or less tomato tomahtoe thing. doing in lapply doesn't far.

lapply( tdatm.sp, function(x){      med=median(x)     mad=mad(x)     x[x>med+3*mad | x<med-3*mad]=na     return(x)} )

now lapply sending anonymous function. if didn't want function hanging around afterwards handy syntax.


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