Question

我有以下data.table：

  dt=structure(list(a = c("10", "10", "20", "30", "10", "25", "10"
    ), b = c("0.605887455840394", "0", "0.709466017509524", "0", 
    "0.585528817843856", "-0.109303314681054", "-0.453497173462763"
    ), c = c("-0.919322002474128", "0", "0.630098551068391", "0", 
    "-1.81795596770373", "-0.276184105225216", "-0.284159743943371"
    ), d = c("-0.750531994502331", "0", "1.81731204370422", "0", 
    "-0.116247806352002", "0.370627864257954", "0.520216457554957"
    ), e = c("0.298723699267293", "0", "-0.886357521243213", "0", 
    "0.816899839520583", "-0.331577589942552", "1.12071265166956"
    ), key = c("A", "A", "B", "B", "C", "C", "C")), .Names = c("a", 
    "b", "c", "d", "e", "key"), row.names = c(NA, -7L), class = c("data.table", 
        "data.frame"), sorted = "key")

这给了我一个类似于下面所示的数据表。

    a                  b                  c                  d                  e key
1: 10  0.605887455840394 -0.919322002474128 -0.750531994502331  0.298723699267293   A
2: 10                  0                  0                  0                  0   A
3: 20  0.709466017509524  0.630098551068391   1.81731204370422 -0.886357521243213   B
4: 30                  0                  0                  0                  0   B
5: 10  0.585528817843856  -1.81795596770373 -0.116247806352002  0.816899839520583   C
6: 25 -0.109303314681054 -0.276184105225216  0.370627864257954 -0.331577589942552   C
7: 10 -0.453497173462763 -0.284159743943371  0.520216457554957   1.12071265166956   C

我想做一个子集操作，删除全部为零的行。

我正在考虑

的内容

dt[!(all(i[2:4) == 0)]但我不确定如何在data.table中实际说明这一点

非常感谢任何帮助。

Answer 1

这似乎是使用not-join的绝佳机会。这需要将密钥设置为您希望在

上进行子集化的列

keys <- names(dt)[2:5]
setkeyv(dt, keys)

 dt[!as.list(rep("0", length(keys)))]

请注意，目前您的键列是字符，这比它们是数字更有效。

Answer 2

1）第一行创建一个逻辑向量，用于选择适当的行，第二行选择它们：

ok <- dt[, ! apply(.SD == 0, 1, all), .SDcols = 2:5]
dt[ok]

2）我们也可以用any来表达，节省一个字符加空格：

ok <- dt[, apply(.SD != 0, 1, any), .SDcols = 2:5]
dt[ok]

3）对于少数列，这甚至更短：

dt[ apply(cbind(b, c, d, e) != 0, 1, any) ]

4）并且对于少数列，这一列更短且更简单

dt[ b != 0 | c != 0 | d != 0 | e != 0 ]

Answer 3

这是一个两步解决方案：

dt[
    !dt[,
        .I[all(sapply(.SD,function(x)x=="0"))]
    ,by=1:nrow(dt),.SDcols=letters[2:5]]$V1
]

产生

    a                  b                  c                  d                  e key
1: 10  0.605887455840394 -0.919322002474128 -0.750531994502331  0.298723699267293   A
2: 20  0.709466017509524  0.630098551068391   1.81731204370422 -0.886357521243213   B
3: 10  0.585528817843856  -1.81795596770373 -0.116247806352002  0.816899839520583   C
4: 25 -0.109303314681054 -0.276184105225216  0.370627864257954 -0.331577589942552   C
5: 10 -0.453497173462763 -0.284159743943371  0.520216457554957   1.12071265166956   C

内部部分选择满足条件的行索引“.I”。外括号通过使用非“！”排除这些行来设置“dt”。操作

使用R data.table和'all（）'来子集数据

3 个答案: