当使用`get()`时,j参数中`.SD`的列顺序不同

2019年7月20日 186次阅读

我经常使用data.table中的.SDcols选项转换数据子集.发送到j的.SD列与原始data.table的顺序相同是有道理的.

已编辑以正确识别问题

很好的.SD列具有与.SDcols参数中指定的顺序相同的顺序.当在j参数中使用get时(至少在lapply调用内),这不会发生.在这种情况下,.SD表列保持其原始顺序.

有没有办法覆盖这种行为？

没有get的例子很好

# library(data.table)
dt = data.table(col1 = rep(LETTERS[1:3], 4), 
            b = rnorm(12),
            a = 1:12,
            c = LETTERS[1:12])

# columns I want to do something to
d.vars = c('a', 'b')  #' names in different order than names(dt)


# Generate columns of first differences by group
dt[, paste('d', d.vars, sep='.') := 
    lapply(.SD, function(L) L  - shift(L, n = 1, type='lag') ), 
      keyby = col1, .SDcols = d.vars]

结果是将差异值分配给“错误”列,因为我的命名向量(d.vars)的排序与dt中的列不同.结果是：

结果与预期的一样,.SD表的列的排序方式与d.vars中的名称相同.

> dt
   col1           b  a c d.a         d.b
1:    A -0.28901751  1 A  NA          NA
2:    A  0.65746901  4 D   3  0.94648651
3:    A -0.10602462  7 G   3 -0.76349362
4:    A -0.38406252 10 J   3 -0.27803790
5:    B -1.06963450  2 B  NA          NA
6:    B  0.35137273  5 E   3  1.42100723
7:    B  0.43394046  8 H   3  0.08256772
8:    B  0.82525042 11 K   3  0.39130996
9:    C  0.50421710  3 C  NA          NA
10:   C -1.09493665  6 F   3 -1.59915375
11:   C -0.04858163  9 I   3  1.04635501
12:   C  0.45867279 12 L   3  0.50725443

这是预期的输出,因为尽管列中的列顺序为d,但是j处理列中的第一个和第二个bb中的lapply.

get的示例表现不同

dt2 = data.table(col1 = rep(LETTERS[1:3], 4),
            b = rnorm(12),
            a = 1:12,
            neg = -1,
            c = LETTERS[1:12])

# columns I want to do something to
d.vars = c('a', 'b')  #' names in different order than names(dt)

# name of variable to be called in j.
negate <- 'neg'

dt2[, paste('d', d.vars, sep='.') :=
    lapply(.SD, function(L)  {(L  - shift(L, n = 1, type='lag') ) * get(negate) }),
       keyby = col1, .SDcols = d.vars]

现在,新创建的列的命名与d.vars中的名称顺序不一致：

 > dt2
    col1          b  a neg c         d.a d.b
 1:    A -0.3539066  1  -1 A          NA  NA
 2:    A  0.2702374  4  -1 D -0.62414408  -3
 3:    A -0.7834941  7  -1 G  1.05373150  -3
 4:    A -1.2765652 10  -1 J  0.49307118  -3
 5:    B -0.2936422  2  -1 B          NA  NA
 6:    B -0.2451996  5  -1 E -0.04844252  -3
 7:    B -1.6577614  8  -1 H  1.41256181  -3
 8:    B  1.0668059 11  -1 K -2.72456737  -3
 9:    C -0.1160938  3  -1 C          NA  NA
10:    C -0.7940771  6  -1 F  0.67798333  -3
11:    C  0.2951743  9  -1 I -1.08925140  -3
12:    C -0.4508854 12  -1 L  0.74605969  -3

在第二个例子中,b列首先由lapply处理,因此分配给d.a.

如果我直接引用neg(即,我不使用get),那么结果是预期的：lapply按照d.vars中给出的顺序处理.SD列.

附：谢谢data.table团队！我喜欢这个包！

最佳答案 根据描述,我们可以使用匹配来匹配’d.vars’和’dt'(‘d.vars1’)的列名,然后使用它来获得正确的顺序

d.vars1 <- d.vars[match(names(dt), d.vars, nomatch = 0)]
dt[, paste0("d.",d.vars1) := lapply(.SD, function(L)
        L  - shift(L, n = 1, type='lag') ), keyby = col1, .SDcols = d.vars1]
dt
#    col1           b  a c         d.b d.a
# 1:    A -0.28901751  1 A          NA  NA
# 2:    A  0.65746901  4 D  0.94648652   3
# 3:    A -0.10602462  7 G -0.76349363   3
# 4:    A -0.38406252 10 J -0.27803790   3
# 5:    B -1.06963450  2 B          NA  NA
# 6:    B  0.35137273  5 E  1.42100723   3
# 7:    B  0.43394046  8 H  0.08256773   3
# 8:    B  0.82525042 11 K  0.39130996   3
# 9:    C  0.50421710  3 C          NA  NA
#10:    C -1.09493665  6 F -1.59915375   3
#11:    C -0.04858163  9 I  1.04635502   3
#12:    C  0.45867279 12 L  0.50725442   3

更新

基于新数据集

d.vars1 <- d.vars[match(names(dt2), d.vars, nomatch = 0)]
dt2[, paste0('d.', d.vars1) := lapply(.SD, function(L) 
    L  - shift(L, n = 1, type='lag') * get(negate) ), 
            keyby = col1, .SDcols = d.vars1]
dt2
#    col1          b  a neg c        d.b d.a
# 1:    A -0.3539066  1  -1 A         NA  NA
# 2:    A  0.2702374  4  -1 D -0.0836692   5
# 3:    A -0.7834941  7  -1 G -0.5132567  11
# 4:    A -1.2765652 10  -1 J -2.0600593  17
# 5:    B -0.2936422  2  -1 B         NA  NA
# 6:    B -0.2451996  5  -1 E -0.5388418   7
# 7:    B -1.6577614  8  -1 H -1.9029610  13
# 8:    B  1.0668059 11  -1 K -0.5909555  19
# 9:    C -0.1160938  3  -1 C         NA  NA
#10:    C -0.7940771  6  -1 F -0.9101709   9
#11:    C  0.2951743  9  -1 I -0.4989028  15
#12:    C -0.4508854 12  -1 L -0.1557111  21
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