:
lapply(myList, function(x) within(x, Init[seq_len(numtake[1])] <- rep("new", numtake[1])))
$TH.9990.1
Init Monthly Dlr Cls.Accoun Cur.Balance meandiff numtake val
12920 new 1 9990 502-****** 3949.85 48.30333 1 TH.9990.1
12921 TH 1 9990 502-****** 4112.69 211.14333 1 TH.9990.1
12922 TH 1 9990 502-****** 3642.10 259.44667 1 TH.9990.1
$VAF.9990.1
Init Monthly Dlr Cls.Accoun Cur.Balance meandiff numtake val
13834 new 1 9990 502-****** 2600.24 78.7925 2 VAF.9990.1
13835 new 1 9990 502-****** 3618.78 939.7475 2 VAF.9990.1
13836 VAF 1 9990 502-****** 4337.44 1658.4075 2 VAF.9990.1
13837 VAF 1 9990 502-****** 159.67 2519.3625 2 VAF.9990.1
numtakeの最初の観察を選択するようにに繰り返される「新しい」割り当ての値だけでなく、引数の数を選択するために使用されますInit変数での置換に使用する要素値。 within
は入力を減らし、変更されたdata.frameおよびlapply
ループをリストに戻します。
データ
structure(list(TH.9990.1 = structure(list(Init = c("TH", "TH",
"TH"), Monthly = c(1L, 1L, 1L), Dlr = c(9990L, 9990L, 9990L),
Cls.Accoun = c("502-******", "502-******", "502-******"),
Cur.Balance = c(3949.85, 4112.69, 3642.1), meandiff = c(48.30333,
211.14333, 259.44667), numtake = c(1L, 1L, 1L), val = c("TH.9990.1",
"TH.9990.1", "TH.9990.1")), .Names = c("Init", "Monthly",
"Dlr", "Cls.Accoun", "Cur.Balance", "meandiff", "numtake", "val"
), row.names = c("12920", "12921", "12922"), class = "data.frame"),
VAF.9990.1 = structure(list(Init = c("VAF", "VAF", "VAF",
"VAF"), Monthly = c(1L, 1L, 1L, 1L), Dlr = c(9990L, 9990L,
9990L, 9990L), Cls.Accoun = c("502-******", "502-******",
"502-******", "502-******"), Cur.Balance = c(2600.24, 3618.78,
4337.44, 159.67), meandiff = c(78.7925, 939.7475, 1658.4075,
2519.3625), numtake = c(2L, 2L, 2L, 2L), val = c("VAF.9990.1",
"VAF.9990.1", "VAF.9990.1", "VAF.9990.1")), .Names = c("Init",
"Monthly", "Dlr", "Cls.Accoun", "Cur.Balance", "meandiff",
"numtake", "val"), row.names = c("13834", "13835", "13836",
"13837"), class = "data.frame")), .Names = c("TH.9990.1",
"VAF.9990.1"))
あなたが何をすべきかについて、より良い方向を受け取るために作られている任意のコードの試みを共有してください。 – JJFord3