从data.frame语法切换到data.table语法对我来说仍然不顺利。我认为以下事情应该是微不足道的，但不是。我在这里做错了什么：
> DT = data.table(x=rep(c("a","b","c"),each=3), y=c(1,3,6), v=1:9)
> DT
x y v
1: a 1 1
2: a 3 2
3: a 6 3
4: b 1 4
5: b 3 5
6: b 6 6
7: c 1 7
8: c 3 8
9: c 6 9
我想要这样的事情：
cols = c("y", "v") # a vector of column names or indexes
DT[rowSums(cols) > 5] # Take only rows where
# values at colums y and v satisfy a condition. 'rowSums' here is just an
# example it can be any function that return TRUE or FALSE when applied
# to values of the row.
这项工作，但如果我想提供动态列名怎么办？我的表有很多列？
>DT[eval( quote(y + v > 5))] #and the following command gives the same result
> DT[y + v > 5]
x y v
1: a 6 3
2: b 3 5
3: b 6 6
4: c 1 7
5: c 3 8
6: c 6 9
> DT[lapply(.SD, sum) > 5, .SDcols = 2:3] # Want the same result as above
Empty data.table (0 rows) of 3 cols: x,y,v
> DT[lapply(.SD, sum) > 5, ,.SDcols = 2:3]
Empty data.table (0 rows) of 3 cols: x,y,v
> DT[lapply(.SD, sum) > 5, , .SDcols = c("y", "v")]
Empty data.table (0 rows) of 3 cols: x,y,v
在答案后更新 事实证明，有很多方法可以做这件事，我想看看哪个想法是最好的表演者。以下是模拟的时序代码：
nr = 1e7
DT = data.table(x=sample(c("a","b","c"),nr, replace= T),
y=sample(2:5, nr, replace = T), v=sample(1:9, nr, T))
threshold = 5
cols = c("y", "v")
col.ids = 2:3
filter.methods = 'DT[DT[, rowSums(.SD[, cols, with = F]) > threshold]]
DT[DT[, rowSums(.SD[, col.ids, with = F]) > threshold]]
DT[DT[, rowSums(.SD) > threshold, .SDcols = cols]]
DT[DT[, rowSums(.SD) > threshold, .SDcols = c("y", "v")]]
DT[DT[, rowSums(.SD) > threshold, .SDcols = col.ids]]
DT[ ,.SD[rowSums(.SD[, col.ids, with = F]) > threshold]]
DT[ ,.SD[rowSums(.SD[, cols, with = F]) > threshold]]
DT[, .SD[rowSums(.SD) > threshold], .SDcols = cols, by = x]
DT[, .SD[rowSums(.SD) > threshold], .SDcols = col.ids, by = x]
DT[, .SD[rowSums(.SD) > threshold], .SDcols = c("y", "v"), by = x]
DT[Reduce(`+`,eval(cols))>threshold]
DT[Reduce(`+`, mget(cols)) > threshold]
'
fm <- strsplit(filter.methods, "\n")
fm <- unlist(fm)
timing = data.frame()
rn = NULL
for (e in sample(fm, length(fm))) {
# Seen some weird pattern with first item in 'fm', so scramble it
rn <- c(rn, e)
if (e == "DT[Reduce(`+`,eval(cols))>threshold]") {
cols = quote(list(y, v))
} else {
cols = c("y", "v")
}
tm <- system.time(eval(parse(text = e)))
timing <- rbind(timing,
data.frame(
as.list(tm[c("user.self", "sys.self", "elapsed")])
)
)
}
rownames(timing) <- rn
timing[order(timing$elapsed),]
### OUTPUT ####
# user.self sys.self elapsed
# DT[Reduce(`+`,eval(cols))>threshold] 0.416 0.168 0.581
# DT[Reduce(`+`, mget(cols)) > threshold] 0.412 0.172 0.582
# DT[DT[, rowSums(.SD) > threshold, .SDcols = cols]] 0.572 0.316 0.889
# DT[DT[, rowSums(.SD) > threshold, .SDcols = col.ids]] 0.568 0.320 0.889
# DT[DT[, rowSums(.SD) > threshold, .SDcols = c("y", "v")]] 0.576 0.316 0.890
# DT[ ,.SD[rowSums(.SD[, col.ids, with = F]) > threshold]] 0.648 0.404 1.052
# DT[DT[, rowSums(.SD[, cols, with = F]) > threshold]] 0.688 0.368 1.052
# DT[DT[, rowSums(.SD[, col.ids, with = F]) > threshold]] 0.612 0.440 1.053
# DT[ ,.SD[rowSums(.SD[, cols, with = F]) > threshold]] 0.692 0.368 1.058
# DT[, .SD[rowSums(.SD) > threshold], .SDcols = c("y", "v"), by = x] 0.800 0.448 1.248
# DT[, .SD[rowSums(.SD) > threshold], .SDcols = col.ids, by = x] 0.836 0.412 1.248
# DT[, .SD[rowSums(.SD) > threshold], .SDcols = cols, by = x] 0.836 0.416 1.249
因此速度方面的冠军是：
DT[Reduce(`+`,eval(cols))>threshold]
DT[Reduce(`+`, mget(cols)) > threshold]
我更喜欢我的mget。而且我认为其他因为调用rowSums而较慢的原因，而Reduce只会有助于形成表达式。真诚地感谢所有给出答案的人。我很难决定选择接受＆＃39;接受＆＃39;回答。 Reduce - 基于sum操作非常具体，而rowSums - 基础是使用任意函数的示例。
答案 0 :(得分：6)
cols = c("y", "v")
尝试
DT[DT[, rowSums(.SD[, cols, with = F]) > 5]]
或者
DT[DT[, rowSums(.SD[, 2:3, with = F]) > 5]]
或者
DT[DT[, rowSums(.SD) > 5, .SDcols = cols]]
或者
DT[DT[, rowSums(.SD) > 5, .SDcols = c("y", "v")]]
或者
DT[DT[, rowSums(.SD) > 5, .SDcols = 2:3]]
或者
DT[ ,.SD[rowSums(.SD[, 2:3, with = F]) > 5]]
或者
DT[ ,.SD[rowSums(.SD[, cols, with = F]) > 5]]
或者
DT[, .SD[rowSums(.SD) > 5], .SDcols = cols, by = x]
或者
DT[, .SD[rowSums(.SD) > 5], .SDcols = 2:3, by = x]
或者
DT[, .SD[rowSums(.SD) > 5], .SDcols = c("y", "v"), by = x]
每个都会产生
# x y v
# 1: a 6 3
# 2: b 3 5
# 3: b 6 6
# 4: c 1 7
# 5: c 3 8
# 6: c 6 9
一些解释：
.SD也是一个data.table对象，可以在DT范围内运行。从而，
此行DT[ ,rowSums(.SD[, cols, with = F]) > 5]将返回一个逻辑向量，指示DT具有y + v > 5的情况。因此，我们将添加另一个DT，以便在DT
当您使用.SDcols时，它会将.SD缩小到这些列。因此，如果您只执行DT[, .SD[rowSums(.SD) > 5], .SDcols = 2:3]之类的操作，则会丢失x列，因此添加了by = x。
使用.SDcols时的另一个选择是返回逻辑向量，然后将其嵌入另一个DT
答案 1 :(得分：4)
这是另一种可能性：
cols <- quote(list(y, v))
DT[Reduce(`+`,eval(cols))>5]
或者，如果您希望将cols保留为字符向量：
cols <- c('y', 'v')
DT[Reduce(`+`, mget(cols)) > 5]
答案 2 :(得分：2)
一种方法是：
cols <- quote(list(y, v))
DT[DT[,Reduce(`+`,eval(cols))>5]]
# x y v
# 1: a 6 3
# 2: b 3 5
# 3: b 6 6
# 4: c 1 7
# 5: c 3 8
# 6: c 6 9