我提供了一个我工作的大型数据框的小例子＆gt; 1,000列＆amp; ＆gt; 200行。我想为每列的每个重复行创建一行而不是两行，同时取与列＃1相关的每个值的平均值（平均列为Col 6-8，依此类推）
Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10
A-001 7 40 1 J 3.985645 4.231623 2.36987 9.36545852 8.369663
A-001 7 40 1 J 1.458798 7.652123 1.236985 4.236987 1.22556633
B-002 8 50 0 K 5.00212 8.369562 7.4569852 5.36692 4.6632121
B-002 8 50 0 K 9.02336 1.2120145 3.0014588 8.214569 5.2223698
C-003 10 60 1 L 1.451203 5.321455 8.25963 2.03369878 4.3336988
C-003 10 60 1 L 1.65653 2.369898 8.2136999 7.21458777 5.3366
D-004 3 70 0 M 5.323211 1.147852 7.20014 5.36989 2.36555
D-004 3 70 0 M 4.36969 5.231478 4.23698 3.645478 9.214563
E-005 4 80 1 N 8.123256 9.2356478 5.3696 4.698889 7.366695
E-005 4 80 1 N 7.9632145 0.004555 1.24789 7.3696969 1.23655
预期输出：
col 1 col 2 col 3 col 4 col 5 col 6
A-001 7 40 1 J 2.49282
B-002 8 50 0 K 7.01274
C-003 10 60 1 L 1.55387
D-004 3 70 0 M 4.84645
E-005 4 80 1 N 8.04324
我提前道歉，因为没有以适当的格式制作 预期感谢提前提供的任何帮助
dd <- structure(list(col1 = structure(c(1L, 1L, 2L, 2L, 3L, 3L, 4L,
4L, 5L, 5L), .Label = c("A-001", "B-002", "C-003", "D-004", "E-005"
), class = "factor"), col2 = structure(c(1L, 1L, 2L, 2L, 3L,
3L, 4L, 4L, 5L, 5L), .Label = c("7", "8", "10", "3", "4"), class = "factor"),
col3 = c(40L, 40L, 50L, 50L, 60L, 60L, 70L, 70L, 80L, 80L
), col4 = c(1, 1, 0, 0, 1, 1, 0, 0, 1, 1), col5 = c(JL, JL, KL, KL, LL, LL, ML, ML, NL,
NL), col6 = c(3.985645, 1.458798, 5.00212, 9.02336, 1.451203, 1.65653, 5.323211,
4.36969, 8.123256, 7.9632145), col7 = c(4.231623, 7.652123, 8.369562, 1.2120145,
5.321455, 2.369898, 1.147852, 5.231478, 9.2356478, 0.004555), col8 = c(2.36987,
1.236985, 7.4569852, 3.0014588, 8.25963, 8.2136999, 7.20014, 4.23698, 5.3696,
1.24789), col9 = c(9.36545852, 4.236987, 5.36692, 8.214569, 2.03369878, 7.21458777,
5.36989, 3.645478, 4.698889, 7.3696969), col10 = c(8.369663, 1.22556633, 4.6632121,
5.2223698, 4.3336988, 5.3366, 2.36555, 9.214563, 7.366695, 1.23655)), .Names =
c("col1", "col2", "col3", "col4", "col5", "col6", "col7", "col8", "col9", "col10"),
class = "data.frame", row.names = c(NA, -10L))
答案 0 :(得分：2)
这使用dplyr，这对于大数据来说会很快。第一部分提取未聚合的唯一列，第二部分执行聚合（通过col1组表示）。然后将这两个部分绑定在一起（bind_cols）。我重命名了你的列以删除空格，如果你留下空格你需要使用带有名字的反引号（即`col 1`）
library(dplyr)
cols <- 6:10 # columns to average
bind_cols(
dat %>% distinct(col1) %>% .[,-cols], # pull out columns we aren't aggregating
dat[,c(1, cols)] %>% group_by(col1) %>%
summarise_each(funs(mean)) %>% .[,-1] # aggregate other columns
)
# col1 col2 col3 col4 col5 col6 col7 col8 col9 col10
# 1 A-001 7 40 1 J 2.49282 5.94187 1.80343 6.80122 4.79762
# 2 B-002 8 50 0 K 7.01274 4.79079 5.22922 6.79075 4.94279
# 3 C-003 10 60 1 L 1.55387 3.84568 8.23667 4.62414 4.83515
# 4 D-004 3 70 0 M 4.84645 3.18967 5.71856 4.50768 5.79006
# 5 E-005 4 80 1 N 8.04324 4.62010 3.30875 6.03429 4.30162
这太复杂了，感谢@StevenBeaupré的简单版本（* _each函数当然可以删除列！）
dat %>% group_by(col1) %>% mutate_each(funs(mean), -(1:5)) %>% distinct
答案 1 :(得分：2)
使用data.table
library(data.table)#v1.9.5+
cols <- 6:10
setDT(dd)[, lapply(.SD, mean), by=c(names(dd)[1:5]), .SDcols=cols]
# col1 col2 col3 col4 col5 col6 col7 col8 col9
#1: A-001 7 40 1 J 2.49282 5.94187 1.80343 6.80122
#2: B-002 8 50 0 K 7.01274 4.79079 5.22922 6.79075
#3: C-003 10 60 1 L 1.55387 3.84568 8.23667 4.62414
#4: D-004 3 70 0 M 4.84645 3.18967 5.71856 4.50768
#5: E-005 4 80 1 N 8.04324 4.62010 3.30875 6.03429
# col10
#1: 4.79762
#2: 4.94279
#3: 4.83515
#4: 5.79006
#5: 4.30162
注意：在提供的示例中，对于每组＆＃39; col1＆＃39;，第2到第5列值是相同的。因此，我们可以使用1：5列作为分组变量。
如果col2：col5中的唯一值的长度对于每个＆＃39; col1＆＃39;是> 1。变量并希望保留每个col1组的col2：col5的第一行
DT1 <- setDT(dd)[, lapply(.SD, mean), by = col1, .SDcols=cols]
setkey(unique(dd[,-cols, with=FALSE], by='col1'),col1)[DT1]
# col1 col2 col3 col4 col5 col6 col7 col8 col9
#1: A-001 7 40 1 J 2.49282 5.94187 1.80343 6.80122
#2: B-002 8 50 0 K 7.01274 4.79079 5.22922 6.79075
#3: C-003 10 60 1 L 1.55387 3.84568 8.23667 4.62414
#4: D-004 3 70 0 M 4.84645 3.18967 5.71856 4.50768
#5: E-005 4 80 1 N 8.04324 4.62010 3.30875 6.03429
# col10
#1: 4.79762
#2: 4.94279
#3: 4.83515
#4: 5.79006
#5: 4.30162