Question

我有一个包含10列的数据表。

town    
tc  
one  
two  
three   
four    
five    
six  
seven   
total

需要为列生成均值＆＃34;一个＆＃34;到＆＃34;总计＆＃34;我正在使用，

DTmean <- DT[,(lapply(.SD,mean)),by = .(town,tc),.SDcols=3:10]

这会生成平均值，但我希望列名称以＆＃34; _mean＆＃34;为后缀。我们应该怎么做？希望前两列保持与＆＃34; town＆＃34;和＆＃34; tc＆＃34;。我尝试了以下但是它重新命名所有＆＃34;一个＆＃34;到＆＃34;总计＆＃34;只是＆＃34; _mean＆＃34;

for (i in 3:10) {
  setnames(DTmean,i,paste0(names(i),"_mean"))
}

Answer 1

如果您想以data.table方式执行此操作，则应使用setnames，如下所示：

setnames(DTmean, 3:10, paste0(names(DT)[3:10], '_mean'))

或：

cols <- names(DT)[3:10]
setnames(DTmean, cols, paste0(cols, '_mean'))

此外，在聚合所有其他列时，您不需要.SDcols语句。因此，使用DT[, lapply(.SD,mean), by = .(town,tc)]可以获得与使用DT[, (lapply(.SD,mean)), by = .(town,tc), .SDcols=3:10]相同的结果。

在以下示例数据集中：

set.seed(71)
DT <- data.table(town = rep(c('A','B'), each=10),
                 tc = rep(c('C','D'), 10),
                 one = rnorm(20,1,1),
                 two = rnorm(20,2,1),
                 three = rnorm(20,3,1),
                 four = rnorm(20,4,1),
                 five = rnorm(20,5,2),
                 six = rnorm(20,6,2),
                 seven = rnorm(20,7,2),
                 total = rnorm(20,28,3))

使用：

DTmean <- DT[, lapply(.SD,mean), by = .(town,tc)]
setnames(DTmean, 3:10, paste0(names(DT)[3:10], '_mean'))

给出：

> DTmean
   town tc  one_mean two_mean three_mean four_mean five_mean six_mean seven_mean total_mean
1:    A  C 1.7368898 1.883586   3.358440  4.849896  4.742609 5.089877   6.792513   29.20286
2:    A  D 0.8906842 1.826135   3.267684  3.760931  6.210145 7.320693   5.571687   26.56142
3:    B  C 1.4037955 2.474836   2.587920  3.719658  3.446612 6.510183   8.309784   27.80012
4:    B  D 0.8103511 1.153000   3.360940  3.945082  5.555999 6.198380   8.652779   28.95180

回复你的评论：如果你想同时计算均值和sd，你可以做（改编自我的回答here）：

DT[, as.list(unlist(lapply(.SD, function(x) list(mean = mean(x), sd = sd(x))))), by = .(town,tc)]

给出：

   town tc  one.mean    one.sd two.mean    two.sd three.mean  three.sd four.mean  four.sd five.mean   five.sd six.mean    six.sd seven.mean seven.sd total.mean total.sd
1:    A  C 0.2981842 0.3556520 1.578174 0.7788545   2.232366 0.9047046  4.896201 1.238877  4.625866 0.7436584 7.607439 1.7262628   7.949366 1.772771   28.94287 3.902602
2:    A  D 1.2099018 1.0205252 1.686068 1.5497989   2.671027 0.8323733  4.811279 1.404794  7.235969 0.7883873 6.765797 2.7719942   6.657298 1.107843   27.42563 3.380785
3:    B  C 0.9238309 0.6679821 2.525485 0.8054734   3.138298 1.0111270  3.876207 0.573342  3.843140 2.1991052 4.942155 0.7784024   6.783383 2.595116   28.95243 1.078307
4:    B  D 0.8843948 0.9384975 1.988908 1.0543981   3.673393 1.3505701  3.957534 1.097837  2.788119 1.9089660 6.463784 0.7642144   6.416487 2.041441   27.88205 3.807119

但是，最好以长格式存储它。为此，您可以使用data.table的{{1}}函数，如下所示：

melt

或更简单的操作：

cols <- names(DT)[3:10]
DT2 <- melt(DT[, as.list(unlist(lapply(.SD, function(x) list(mn = mean(x), sdev = sd(x))))), by = .(town,tc)], 
            id.vars = c('town','tc'), 
            measure.vars = patterns('.mn','.sdev'),
            value.name = c('mn','sdev'))[, variable := cols[variable]]

导致：

DT2 <- melt(DT, id.vars = c('town','tc'))[, .(mn = mean(value), sdev = sd(value)), by = .(town,tc,variable)]

在回复您的上一条评论时，您可以按如下方式检测异常值：

> DT2
    town tc variable         mn      sdev
 1:    A  C      one  0.2981842 0.3556520
 2:    A  D      one  1.2099018 1.0205252
 3:    B  C      one  0.9238309 0.6679821
 4:    B  D      one  0.8843948 0.9384975
 5:    A  C      two  1.5781743 0.7788545
 6:    A  D      two  1.6860675 1.5497989
 7:    B  C      two  2.5254855 0.8054734
 8:    B  D      two  1.9889082 1.0543981
 9:    A  C    three  2.2323655 0.9047046
10:    A  D    three  2.6710267 0.8323733
11:    B  C    three  3.1382982 1.0111270
12:    B  D    three  3.6733929 1.3505701
.....

给出：

DT3 <- melt(DT, id.vars = c('town','tc'))
DT3[, `:=` (mn = mean(value), sdev = sd(value)), by = .(town,tc,variable)
    ][, outlier := +(value < mn - sdev | value > mn + sdev)]

如何更改数据表中的几个列名称

1 个答案: