按组汇总并获得不同数据的非NA值的计数，均值和sd.frame列

添加表格 这是我拥有的数据 这是我从代码中获得的数据 这是我想要的表格 您会注意到，在具有"数据框中，按治疗组对var1列中不存在的行进行计数会得出以下结果: veh-9图4-83-102-5 但是当使用sum(！is.na(x)时，我得到以下内容 veh-6图4-53-102-5 我认为这是因为该函数同时使用var1和var2来求和非缺失数.我不知道该如何纠正.最好杰克解决方案这是一种data.table方法: 数据 您拥有的数据难以读入R中-请使用dput()等使其他人更容易使用> dput(dt)structure(list(someting = c("503", "553", "599", "647", "695", "728", "760", "793", "826", "859", "907", "955", "1003", "1036", "1084", "1131", "1179", "1226", "1274", "1322", "1355", "1402", "1450", "1497", "1545"), treatment = c("gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.4", "gr.4"), var1 = c(8, NA, 3, 3, NA, NA, NA, NA, NA, 8, 8, 8, NA, 8, 8, 8, 8, 8, 8, NA, 8, 8, 8, 8, NA), var2 = c(8L, 8L, 8L, 8L, NA, NA, NA, NA, NA, 8L, 8L, 8L, NA, 8L, 8L, 8L, 8L, 8L, 8L, NA, 8L, 8L, 8L, 8L, NA)), .Names = c("someting", "treatment", "var1", "var2"), row.names = c(NA, -25L), class = c("data.table", "data.frame")) 代码 dt[, .(var1.n = sum(!is.na(var1)), var2.n = sum(!is.na(var1)), var1.mean = mean(var1, na.rm = T), var2.mean = mean(var2, na.rm = T)), by = .(treatment)] 输出 treatment var1.n var2.n var1.mean var2.mean1: gr.2 5 5 6 82: gr.3 10 10 8 83: gr.4 1 1 8 8由于某些原因，未读入"veh"条目.因此，输出略有不同，但原理应明确. I am having some difficulty counting non-missing values by group through the function below (which also gives sd, and mean):test <- do.call(data.frame, aggregate(. ~ treatment, have, function(x) c(n = sum(!is.na(x)), mean = mean(x), sd = sd(x))))It ends up giving me the number of non-missing for all columns in the dataframe instead of just a single column.I have been looking through SO for some advice and found this, this, and this helpful, but I can't figure out why the aggregate with the function(x) would combine some columns for the sum(!is.na(x), but not for the mean or sd.EDIT: Adding tablesThis is the data I haveThis is the data I get from my codeThis is the table I wantYou will notice in the 'have' dataframe that counting the non-mising rows in column var1 by treatment group gives the following:veh - 9gr.4 - 8gr.3 - 10gr.2 - 5But when using the sum(!is.na(x) I get the followingveh - 6gr.4 - 5gr.3 - 10gr.2 - 5I believe this is because the function is using both var1 and var2 to sum the number of non-missing. I do not know how to correct for this.Best,Jack 解决方案 Here's a data.table approach: DATAThe data you have is cumbersome to read into R - please use dput() etc. to make it easier for others:> dput(dt)structure(list(someting = c("503", "553", "599", "647", "695", "728", "760", "793", "826", "859", "907", "955", "1003", "1036", "1084", "1131", "1179", "1226", "1274", "1322", "1355", "1402", "1450", "1497", "1545"), treatment = c("gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.2", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.3", "gr.4", "gr.4"), var1 = c(8, NA, 3, 3, NA, NA, NA, NA, NA, 8, 8, 8, NA, 8, 8, 8, 8, 8, 8, NA, 8, 8, 8, 8, NA), var2 = c(8L, 8L, 8L, 8L, NA, NA, NA, NA, NA, 8L, 8L, 8L, NA, 8L, 8L, 8L, 8L, 8L, 8L, NA, 8L, 8L, 8L, 8L, NA)), .Names = c("someting", "treatment", "var1", "var2"), row.names = c(NA, -25L), class = c("data.table", "data.frame"))CODEdt[, .(var1.n = sum(!is.na(var1)), var2.n = sum(!is.na(var1)), var1.mean = mean(var1, na.rm = T), var2.mean = mean(var2, na.rm = T)), by = .(treatment)]OUTPUT treatment var1.n var2.n var1.mean var2.mean1: gr.2 5 5 6 82: gr.3 10 10 8 83: gr.4 1 1 8 8For some reason the "veh" entries weren't read in. Hence the output is slightly different but the principle ought to be clear. 这篇关于按组汇总并获得不同数据的非NA值的计数，均值和sd.frame列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！