搜索了类似的帖子后，我发布了我的问题。每个站点我有几年的月降雨量变量。我需要计算多年来的月平均降雨量。我给出了一个简单的数据框如下。我需要创建一个新的数据框架，其中包含每个站点的月平均值（12）。
d<-structure(list(ID = structure(1:4, .Label = c("A", "B", "C",
"D"), class = "factor"), X2000_1 = c(25L, 42L, 74L, 52L), X2000_2 = c(15L,
15L, 51L, 12L), X2000_3 = c(14L, 21L, 25L, 41L), X2000_4 = c(74L,
4L, 23L, 51L), X2000_5 = c(15L, 25L, 65L, 12L), X2000_6 = c(31L,
23L, 15L, 25L), X2001_1 = c(52L, 54L, 18L, 63L), X2001_2 = c(85L,
165L, 12L, 12L), X2001_3 = c(25L, 36L, 20L, 14L), X2001_4 = c(1L,
17L, 23L, 52L), X2001_5 = c(24L, 45L, 12L, 15L), X2001_6 = c(3L,
23L, 45L, 52L)), .Names = c("ID", "X2000_1", "X2000_2", "X2000_3",
"X2000_4", "X2000_5", "X2000_6", "X2001_1", "X2001_2", "X2001_3",
"X2001_4", "X2001_5", "X2001_6"), class = "data.frame", row.names = c(NA,
-4L))
输出应该是;
df<-data.frame(id = c("A","B","C","D"))
df[c("jan","feb","mar","apr","may","jun")]<-NA
例如，单元格A1应包含X2000_1和X2001_1的平均降雨量
我尝试了下面的代码，但它不起作用可能是因为我正在使用数据框。任何帮助将不胜感激。
n = 6
unname(tapply(d, (seq_along(d)-1) %/% n, sum))
我的实际数据框的列名是
c("est", "X1990_1", "X1990_2", "X1990_3", "X1990_4", "X1990_5",
"X1990_6", "X1990_7", "X1990_8", "X1990_9", "X1990_10", "X1990_11",
"X1990_12", "X1991_1", "X1991_2", "X1991_3", "X1991_4", "X1991_5",
"X1991_6", "X1991_7", "X1991_8", "X1991_9", "X1991_10", "X1991_11",
"X1991_12", "X1992_1", "X1992_2", "X1992_3", "X1992_4", "X1992_5",
"X1992_6", "X1992_7", "X1992_8", "X1992_9", "X1992_10", "X1992_11",
"X1992_12", "X1993_1", "X1993_2", "X1993_3", "X1993_4", "X1993_5",
"X1993_6", "X1993_7", "X1993_8", "X1993_9", "X1993_10", "X1993_11",
"X1993_12", "X1994_1", "X1994_2", "X1994_3", "X1994_4", "X1994_5",
"X1994_6", "X1994_7", "X1994_8", "X1994_9", "X1994_10", "X1994_11",
"X1994_12", "X1995_1", "X1995_2", "X1995_3", "X1995_4", "X1995_5",
"X1995_6", "X1995_7", "X1995_8", "X1995_9", "X1995_10", "X1995_11",
"X1995_12", "X1996_1", "X1996_2", "X1996_3", "X1996_4", "X1996_5",
"X1996_6", "X1996_7", "X1996_8", "X1996_9", "X1996_10", "X1996_11",
"X1996_12", "X1997_1", "X1997_2", "X1997_3", "X1997_4", "X1997_5",
"X1997_6", "X1997_7", "X1997_8", "X1997_9", "X1997_10", "X1997_11",
"X1997_12", "X1998_1", "X1998_2", "X1998_3", "X1998_4", "X1998_5",
"X1998_6", "X1998_7", "X1998_8", "X1998_9", "X1998_10", "X1998_11",
"X1998_12", "X1999_1", "X1999_2", "X1999_3", "X1999_4", "X1999_5",
"X1999_6", "X1999_7", "X1999_8", "X1999_9", "X1999_10", "X1999_11",
"X1999_12", "X2000_1", "X2000_2", "X2000_3", "X2000_4", "X2000_5",
"X2000_6", "X2000_7", "X2000_8", "X2000_9", "X2000_10", "X2000_11",
"X2000_12")
答案 0 :(得分：2)
您可以从列名称中提取月份作为变量，并按月份变量将数据帧拆分为列表，并使用rowMeans()函数计算每个子数据帧的行平均值：
# extract the months for each column
mon <- sub(".*_(\\d+)$", "\\1", names(d)[-1])
# split the data frame by columns and calculate the rowMeans
cbind.data.frame(d[1], lapply(split.default(d[-1], mon), rowMeans))
# ID 1 2 3 4 5 6
#1 A 38.5 50.0 19.5 37.5 19.5 17.0
#2 B 48.0 90.0 28.5 10.5 35.0 23.0
#3 C 46.0 31.5 22.5 23.0 38.5 30.0
#4 D 57.5 12.0 27.5 51.5 13.5 38.5
答案 1 :(得分：2)
您也可以使用一些reshape来处理长数据集以及制表：
tmp <- reshape(d, idvar="ID", sep="_", direction="long", varying=-1)
xtabs(rowMeans(cbind(X2000,X2001)) ~ ID + time, data=tmp)
# time
#ID 1 2 3 4 5 6
# A 38.5 50.0 19.5 37.5 19.5 17.0
# B 48.0 90.0 28.5 10.5 35.0 23.0
# C 46.0 31.5 22.5 23.0 38.5 30.0
# D 57.5 12.0 27.5 51.5 13.5 38.5
答案 2 :(得分：2)
以下是使用Reduce和+
cbind(d[1], Reduce(`+`, list(d[2:7], d[8:13]))/2)
# ID X2000_1 X2000_2 X2000_3 X2000_4 X2000_5 X2000_6
#1 A 38.5 50.0 19.5 37.5 19.5 17.0
#2 B 48.0 90.0 28.5 10.5 35.0 23.0
#3 C 46.0 31.5 22.5 23.0 38.5 30.0
#4 D 57.5 12.0 27.5 51.5 13.5 38.5
或者只是
cbind(d[1], (d[2:7] + d[8:13])/2)
答案 3 :(得分：1)
假设我们将第一列作为ID，并且所有列均匀分布。
我们可以将数据帧分成两半并得到它们之间的平均值。
cbind(d[1],(d[2:ceiling(ncol(d)/2)] + d[(ceiling(ncol(d)/2) + 1):ncol(d)])/2)
# ID X2000_1 X2000_2 X2000_3 X2000_4 X2000_5 X2000_6
#1 A 38.5 50.0 19.5 37.5 19.5 17.0
#2 B 48.0 90.0 28.5 10.5 35.0 23.0
#3 C 46.0 31.5 22.5 23.0 38.5 30.0
#4 D 57.5 12.0 27.5 51.5 13.5 38.5
显然，我们总是可以通过对列号进行硬编码来实现。
cbind(d[1], (d[2:7] + d[8:13])/2)
然而，上面提到的方法是一般化的，即使我们有超过13列也会有效。
答案 4 :(得分：0)
据我所知，要获取文件的签出信息，您需要找出工作区，然后找到这些工作区的所有待定更改。