這是一個新手問題,但是,我發現很難理解如何正確使用lapply,特別是當使用的ID不是數字時。使用lapply的數據集摘要
有可能更好的方法來試圖找到我想到的總結,但現在,我試圖使用lapply。基本上,我有一個17列大DF。兩列是ID和日期。並非所有的ID在給定的列名稱中都有記錄的值。我感興趣的是查找每列可用的總行數,以及該列存在的唯一ID的數量。我有一個讓事情變得更清晰的例子。例如,Var8只有6行數據可用,因此它有6個唯一的ID。另外,Var15有20行和12個唯一的ID。但我想知道所有Var15的情況。我能做到這一點使用
Var8=df[!(is.na(df$Var8)),]
length(df$ID)
length(unique(df$ID))
remove(Var8)
但是,試圖將自動手動:
lapply(COL.NAMES, function(x){
temp=df[!(is.na(df$paste(x))),]
rows=length(temp$ID)
num_comp=length(unique(temp$ID))
return(rows)
return(num_comp)
remove(temp)
})
給我留下了一個錯誤:試圖將非功能。
COL.NAMES<-c("Var1","Var2","Var3","Var4","Var5","Var6","Var7","Var8","Var9","Var10","Var11","Var12","Var13","Var14","Var15")
structure(list(ID = structure(c(1L, 5L, 6L, 7L, 8L, 9L, 10L,
11L, 12L, 2L, 3L, 4L, 1L, 5L, 6L, 7L, 8L, 9L, 10L, 11L), .Label = c("Comp1",
"Comp10", "Comp11", "Comp12", "Comp2", "Comp3", "Comp4", "Comp5",
"Comp6", "Comp7", "Comp8", "Comp9"), class = "factor"), Date = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L), .Label = c("0/1/2014", "0/1/2015"), class = "factor"),
Var1 = c(0.57, 0.34, 0.38, 0.93, 0.54, 0.17, 0.08, 0.28,
0.99, 1, 0.61, 0.73, 0.15, 0.09, 0.64, 0.3, 0.12, 0.79, 0.79,
0.15), Var2 = c(0.7, 0.77, 0.93, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.46, 0.26), Var3 = c(0.65,
0.7, 0.83, 0.7, 0.43, 0.81, 0.21, 0.44, 0.25, 0.77, 0.24,
0.29, 0.87, 0.42, 1, NA, NA, NA, NA, 0.79), Var4 = c(1, 0.7,
0.69, NA, NA, NA, NA, 0.2, 0.61, 0.89, 0.45, 0.02, 0.97,
0.33, 0.34, 0.81, 0.99, 0.35, 0.48, 0.33), Var5 = c(0.47,
0.95, 0.38, 0.69, 0.84, 0.21, 0.62, 0.59, 0.45, 0.63, 0.18,
0.49, NA, NA, NA, NA, 0.17, 0.15, 0.6, 0.44), Var6 = c(NA,
NA, NA, NA, 0.24, 0.07, 0.75, 0.24, 0.82, 0.14, 0.86, 0.63,
0.82, 0.92, 0.55, 0.22, 0.87, 0.69, 0.64, 0.73), Var7 = c(0.2,
0.11, 0.82, 0.31, 0.97, NA, NA, NA, NA, 0.83, 0.84, 0.81,
0.72, 0.36, 0.09, 0.15, 0.46, 0.79, 0.75, 0.39), Var8 = c(0.28,
0.55, NA, NA, NA, NA, 0.56, 0.89, 0.92, 0.46, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA), Var9 = c(0.11, 0.36, 1, 0.44,
0.53, 0.6, 0.24, 0.56, 0.6, 0.55, 0.55, 0.05, 0.77, 0.9,
NA, NA, NA, NA, 0.4, 0.33), Var10 = c(0.74, 0.13, 0.09, 0.61,
NA, NA, NA, NA, 0.27, 0.71, 0.56, 0.3, 0.36, 0.44, 0.78,
0.9, 0.46, 0.49, 0.87, 0.36), Var11 = c(0.58, 0.99, 0.07,
0.83, 0.45, 0.07, 0.16, 0.43, 0.34, 0.31, 0.06, 0.67, 0.02,
0.52, 0.19, 0.49, 0.31, 0.02, 0.62, 0.21), Var12 = c(0.93,
0.26, 0.77, 0.8, 0.67, 0.83, 0.12, 0.39, 0.78, 0.75, 0.44,
NA, NA, NA, NA, 0.42, 0.49, 0.06, 0.8, 0.54), Var13 = c(0.44,
0.75, NA, NA, NA, NA, 0.58, 0.3, 0.47, 0.88, 0.36, 0.21,
0.87, 0.33, 0.12, 0.31, 0.95, 0.59, 0.18, 0.43), Var14 = c(0.55,
0.03, 0.37, 0.66, NA, 0.91, 0.78, 0.84, 0.96, 0.34, 0.25,
0.92, 0.71, 0.41, 0.23, 0.54, 0.8, 0.87, 0.3, 0.37), Var15 = c(0.71,
0.66, 0.01, 0.7, 0.4, 0.04, 0.3, 1, 0.59, 0.69, 0.88, 0.28,
0.44, 0.51, 0.2, 0.17, 0.6, 0.11, 0.85, 0.04)), .Names = c("ID",
"Date", "Var1", "Var2", "Var3", "Var4", "Var5", "Var6", "Var7",
"Var8", "Var9", "Var10", "Var11", "Var12", "Var13", "Var14",
"Var15"), class = "data.frame", row.names = c(NA, -20L))
您能不能告訴預期的輸出? – Sotos
Adam Quek下面的答案是你需要的,但是你得到錯誤信息的原因是你的函數中有兩個返回語句。您可以返回一個列表,也可以返回兩個值的向量。你也不需要刪除。 – aichao