我想找到最有效的方法来组合两个数据帧并平均具有不同row.names的列中的值.所以,我想从两个数据中取jsut重叠的row.names并将它们合并为一个.列的值应按平均值进行平均值.示例数据:
mtcars <-
structure(list(mpg = c(21, 21, 22.8, 21.4, 18.7, 18.1, 14.3,
24.4, 22.8, 19.2, 17.8, 16.4, 17.3, 15.2, 10.4, 10.4, 14.7, 32.4,
30.4, 33.9, 21.5, 15.5, 15.2, 13.3, 19.2, 27.3, 26, 30.4, 15.8,
19.7, 15, 21.4), cyl = c(6, 6, 4, 6, 8, 6, 8, 4, 4, 6, 6, 8,
8, 8, 8, 8, 8, 4, 4, 4, 4, 8, 8, 8, 8, 4, 4, 4, 8, 6, 8, 4),
disp = c(160, 160, 108, 258, 360, 225, 360, 146.7, 140.8,
167.6, 167.6, 275.8, 275.8, 275.8, 472, 460, 440, 78.7, 75.7,
71.1, 120.1, 318, 304, 350, 400, 79, 120.3, 95.1, 351, 145,
301, 121), hp = c(110, 110, 93, 110, 175, 105, 245, 62, 95,
123, 123, 180, 180, 180, 205, 215, 230, 66, 52, 65, 97, 150,
150, 245, 175, 66, 91, 113, 264, 175, 335, 109), drat = c(3.9,
3.9, 3.85, 3.08, 3.15, 2.76, 3.21, 3.69, 3.92, 3.92, 3.92,
3.07, 3.07, 3.07, 2.93, 3, 3.23, 4.08, 4.93, 4.22, 3.7, 2.76,
3.15, 3.73, 3.08, 4.08, 4.43, 3.77, 4.22, 3.62, 3.54, 4.11
), wt = c(2.62, 2.875, 2.32, 3.215, 3.44, 3.46, 3.57, 3.19,
3.15, 3.44, 3.44, 4.07, 3.73, 3.78, 5.25, 5.424, 5.345, 2.2,
1.615, 1.835, 2.465, 3.52, 3.435, 3.84, 3.845, 1.935, 2.14,
1.513, 3.17, 2.77, 3.57, 2.78), qsec = c(16.46, 17.02, 18.61,
19.44, 17.02, 20.22, 15.84, 20, 22.9, 18.3, 18.9, 17.4, 17.6,
18, 17.98, 17.82, 17.42, 19.47, 18.52, 19.9, 20.01, 16.87,
17.3, 15.41, 17.05, 18.9, 16.7, 16.9, 14.5, 15.5, 14.6, 18.6
), vs = c(0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0,
0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1), am = c(1,
1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1,
0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1), gear = c(4, 4, 4, 3,
3, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3,
3, 3, 4, 5, 5, 5, 5, 5, 4), carb = c(4, 4, 1, 1, 2, 1, 4,
2, 2, 4, 4, 3, 3, 3, 4, 4, 4, 1, 2, 1, 1, 2, 2, 4, 2, 1,
2, 2, 4, 6, 8, 2)), .Names = c("mpg", "cyl", "disp", "hp",
"drat", "wt", "qsec", "vs", "am", "gear", "carb"), row.names = c("Mazda RX4",
"Mazda RX4 Wag", "Datsun 710", "Hornet 4 Drive", "Hornet Sportabout",
"Valiant", "Duster 360", "Merc 240D", "Merc 230", "Merc 280",
"Merc 280C", "Merc 450SE", "Merc 450SL", "Merc 450SLC", "Cadillac Fleetwood",
"Lincoln Continental", "Chrysler Imperial", "Fiat 128", "Honda Civic",
"Toyota Corolla", "Toyota Corona", "Dodge Challenger", "AMC Javelin",
"Camaro Z28", "Pontiac Firebird", "Fiat X1-9", "Porsche 914-2",
"Lotus Europa", "Ford Pantera L", "Ferrari Dino", "Maserati Bora",
"Volvo 142E"), class = "data.frame")
第二个数据:
mtcars11 <-
structure(list(mpg = c(21, 21, 22.8, 21.4, 18.7, 18.1, 14.3,
24.4, 22.8, 19.2, 17.8, 16.4, 17.3, 15.2, 10.4, 10.4, 14.7, 32.4,
30.4, 33.9, 21.5, 15.5, 15.2, 13.3, 19.2, 27.3, 26, 30.4, 15.8,
19.7), cyl = c(6, 6, 4, 6, 8, 6, 8, 4, 4, 6, 6, 8, 8, 8, 8, 8,
8, 4, 4, 4, 4, 8, 8, 8, 8, 4, 4, 4, 8, 6), disp = c(160, 160,
108, 258, 360, 225, 360, 146.7, 140.8, 167.6, 167.6, 275.8, 275.8,
275.8, 472, 460, 440, 78.7, 75.7, 71.1, 120.1, 318, 304, 350,
400, 79, 120.3, 95.1, 351, 145), hp = c(110, 110, 93, 110, 175,
105, 245, 62, 95, 123, 123, 180, 180, 180, 205, 215, 230, 66,
52, 65, 97, 150, 150, 245, 175, 66, 91, 113, 264, 175), drat = c(3.9,
3.9, 3.85, 3.08, 3.15, 2.76, 3.21, 3.69, 3.92, 3.92, 3.92, 3.07,
3.07, 3.07, 2.93, 3, 3.23, 4.08, 4.93, 4.22, 3.7, 2.76, 3.15,
3.73, 3.08, 4.08, 4.43, 3.77, 4.22, 3.62), wt = c(2.62, 2.875,
2.32, 3.215, 3.44, 3.46, 3.57, 3.19, 3.15, 3.44, 3.44, 4.07,
3.73, 3.78, 5.25, 5.424, 5.345, 2.2, 1.615, 1.835, 2.465, 3.52,
3.435, 3.84, 3.845, 1.935, 2.14, 1.513, 3.17, 2.77), qsec = c(16.46,
17.02, 18.61, 19.44, 17.02, 20.22, 15.84, 20, 22.9, 18.3, 18.9,
17.4, 17.6, 18, 17.98, 17.82, 17.42, 19.47, 18.52, 19.9, 20.01,
16.87, 17.3, 15.41, 17.05, 18.9, 16.7, 16.9, 14.5, 15.5), vs = c(0,
0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0,
0, 0, 0, 1, 0, 1, 0, 0), am = c(1, 1, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1),
gear = c(4, 4, 4, 3, 3, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3,
3, 4, 4, 4, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5), carb = c(4, 4,
1, 1, 2, 1, 4, 2, 2, 4, 4, 3, 3, 3, 4, 4, 4, 1, 2, 1, 1,
2, 2, 4, 2, 1, 2, 2, 4, 6)), .Names = c("mpg", "cyl", "disp",
"hp", "drat", "wt", "qsec", "vs", "am", "gear", "carb"), row.names = c("Mazda RX4",
"Chrysler", "Datsun 710", "Hornet 4 Drive", "Hornet Sportabout",
"Valiant", "Duster 360", "Merc 240D", "Merc 230", "Merc 280",
"Merc 280C", "Merc 450SE", "Nexia", "Merc 450SLC", "Cadillac Fleetwood",
"Lincoln Continental", "Chrysler Imperial", "Fiat 128", "Honda Civic",
"Toyota Corolla", "Toyota Corona", "Dodge Challenger", "AMC Javelin",
"Camaro Z28", "Pontiac Firebirda", "Punto", "Porsche 914-2",
"Lotus Europa", "Ford Pantera T", "Ferrari Dino"), class = "data.frame")
因此,我想到的解决方案是(长期的):
vec_names_mt <- row.names(mtcars) ## so we the row.names from first data
vec_names_mt11 <- row.names(mtcars11) ## so we the row.names from second data
vec_inter <- intersect(vec_names_mt, vec_names_mt11) ## find overlapping names
data_mt <- mtcars[row.names(mtcars) %in% vec_inter, ] ## take the rows from first data which overlaps
data_mt11 <- mtcars11[row.names(mtcars11) %in% vec_inter, ] ## take the rows from second data which overlaps
我们如何将它们结合起来并平均值？知道如何以最简单的方式做到这一点吗？
假设d1并且d2是您的data.frames,这是我如何处理它.你将不得不使用data.table(v1.9.5)的开发版本,虽然,对于mget工作.
require(data.table) # v1.9.5
setkey(setDT(d1, keep.rownames=TRUE), rn)
setkey(setDT(d2, keep.rownames=TRUE), rn)
xcols = names(d1)[-1L]
icols = paste("i.", xcols, sep="")
foo <- function(a, b) mean(c(a, b), na.rm=TRUE)
d1[d2, Map(foo, mget(xcols), mget(icols)), by=.EACHI, nomatch=0L]
我们首先将转换data.frames到data.tables通过参考使用setDT,并转换row names到新的一列(它会自动被命名rn),并设置该列键.
setkey()按指定的列重新排序data.table,并将这些列标记为键列,这将帮助我们执行连接(在这些键列上).
在data.tables中,连接可以通过使用x[i]符号和merge()函数(实现了data.table方法)来完成,但x[i]功能更强大,更灵活.语法x[i]将每一行连接i到x(在键列上)的匹配行.
因此,d1[d2]将返回d2匹配行中的每一行d1,以及其中的所有其他列d2.
d1[d2, nomatch=0L]是内连接的等价物,只返回匹配的行.
d1[d2, Map(foo, mget(xcols), mget(icols)), by=.EACHI, nomatch=0L]为j= Map(...)中的每一行计算表达式d2- 因此by = .EACHI.
总而言之,对于每一行d2,找到匹配的行d1.在提取指定的列xcols,并icols只为那个匹配行,并应用功能foo()将串联的载体,并采取他们mean().并为d2(by = .EACHI)的每一行执行此操作.忽略其中的行在键列()上d2没有任何匹配d1项nomatch=0L.
希望这可以帮助.
看起来您正在寻找行名称之间的两个数据集之间的" 内部联接 ".我建议尝试data.table用于合并以及后来的熔化和压铸操作.
首先,我将重命名mtcars为mtcars2因为mtcars是一个存储的数据集,我不希望它们都覆盖它,因为setDT实际上不能覆盖存储的数据集,所以我们可以说在现实生活中你的数据被调用mtcars2
library(data.table)
mtcars2 <- copy(mtcars)
接下来,我们将转换为data.table对象,同时保留行名称,并为更快的连接设置键
setkey(setDT(mtcars2, keep.rownames = TRUE), rn)
setkey(setDT(mtcars11, keep.rownames = TRUE), rn)
现在,我们将执行一个内连接了rn(关键),同时保持原始列名使用suffixes = NULL
Res <- merge(mtcars2, mtcars11, suffixes = NULL)
现在我们可以melt通过独特的列rn然后dcast计算平均值
dcast(melt(Res, "rn"), rn ~ variable, mean.default)
# rn mpg cyl disp hp drat wt qsec vs am gear carb
# 1: AMC Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
# 2: Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
# 3: Camaro Z28 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
# 4: Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
# 5: Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
# 6: Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
# 7: Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
# 8: Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
# 9: Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
# 10: Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
# 11: Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
# 12: Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
# 13: Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
# 14: Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
# 15: Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
# 16: Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
# 17: Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
# 18: Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
# 19: Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
# 20: Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
# 21: Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
# 22: Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
# 23: Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
# 24: Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
# 25: Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1