獲取每個人的每第n列的總和，並在r中創建新的數據框

Question

搜索了類似的帖子后，我發布了我的問題。 每個站點我有幾年的月降雨量變量。 我需要計算多年來的月平均降雨量。 我給出了一個簡單的數據框如下。 我需要創建一個新的數據框架，其中包含每個站點的月平均值（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))

輸出應該是這樣的;

df<-data.frame(id = c("A","B","C","D"))
df[c("jan","feb","mar","apr","may","jun")]<-NA

例如，單元格A1應包含X2000_1和X2001_1的平均降雨量

我嘗試了我的代碼如下，但它不起作用可能是因為我使用數據框。 任何幫助將非常感激。

n = 6
unname(tapply(d, (seq_along(d)-1) %/% n, sum))

我的實際數據框的列名是

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")

Answer 1

您可以從列名稱中提取月份作為變量，並通過months變量將數據幀拆分為列表，並使用rowMeans()函數計算每個子數據幀的行平均值：

# 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

Answer 2

您還可以通過對一個長數據集進行reshape整形以及制表來實現：

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

Answer 3

這是使用Reduce with +的選項

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

要不就

cbind(d[1], (d[2:7] + d[8:13])/2)

Answer 4

假設，我們將第一列作為ID並且所有列均勻分布。

我們可以將數據幀分成兩半並得到它們之間的平均值。

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

顯然，我們總是可以通過硬編碼列號來完成。

cbind(d[1], (d[2:7] + d[8:13])/2)

但是，上面提到的方法是一般化的，即使我們有超過13列，它也會工作。

Answer 5

據我所知，要獲取文件的簽出信息，您需要找出工作區，然后找到這些工作空間上所有待處理的更改。