R：将列乘以数据表中的常量 [英] R: Multiplying columns by a constant in a data table

问题描述

我尝试更正我的资料表，因此我的资料栏有相同的单位。这是我的一个例子。

I am trying to correct my data table so my columns have the same units. Here's an example of what I have.

hh:mm   A   V   W   kA  V   kW  A   kV  kW
11:00   13.84   470.16  6509.88 14.89   467.85  6964.38 15.74   464.01  7303.13
11:05   12.54   475.17  5959.22 13.40   474.52  6358.89 13.34   473.13  6311.80
11:10   9.73    476.20  4632.14 10.36   473.38  4905.86 10.38   472.73  4907.14
11:15   9.20    479.30  4410.89 9.65    482.79  4659.67 9.73    479.09  4659.33
11:20   11.28   482.22  5437.78 12.03   484.95  5835.33 12.24   476.36  5829.44
11:25   11.66   481.64  5614.56 12.76   479.95  6124.56 12.88   476.86  6139.33
11:30   10.38   475.13  4934.00 11.99   480.96  5760.44 11.50   478.77  5515.13

正如你所看到的，有些列是A，V或W，而其他是kA，kV或kW。我想要做的是通过将kA，kV和kW列乘以1000来将所有这些更改为相同的单位。这是我想要得到的

As you can see, some columns are in "A","V", or "W", while others are in "kA","kV" or "kW". What I am trying to do is to change all these to the same units by multiplying the "kA", "kV" and "kW" columns by 1000. Here's what I want to get

hh:mm   A   V   W   A   V   W   A   V   W
11:00   13.84   470.16  6509.88 14890   467.85  6964380 15.74   464010  7303130
11:05   12.54   475.17  5959.22 13400   474.52  6358890 13.34   473130  6311800
11:10   9.73    476.20  4632.14 10360   473.38  4905860 10.38   472730  4907140
11:15   9.20    479.30  4410.89 9650    482.79  4659670 9.73    479090  4659330
11:20   11.28   482.22  5437.78 12030   484.95  5835330 12.24   476360  5829440
11:25   11.66   481.64  5614.56 12760   479.95  6124560 12.88   476860  6139330
11:30   10.38   475.13  4934.00 11990   480.96  5760440 11.50   478770  5515130

我试过这样做：

units<-which(names(dt)=="kA") # Gives me a vector with the positions needed.
dt[,units:=units*1000] #Multiplies the vector by 1000
names(dt) <- gsub("kA", "A", names(dt)) # Changes "kA" to "A"

units<-which(names(dt)=="kV") # Gives me a vector with the positions needed.
dt[,units:=units*1000] #Multiplies the vector by 1000
names(dt) <- gsub("kV", "V", names(dt)) # Changes "kV" to "V"

units<-which(names(dt)=="kW") # Gives me a vector with the positions needed.
dt[,units:=units*1000] #Multiplies the vector by 1000
names(dt) <- gsub("kW", "W", names(dt)) # Changes "kW" to "W"

在我的第二行，我收到此错误：

On my second line, I receive this error:

Warning message:
In `[.data.table`(x2, , `:=`(units, units * 1000)) :
  Supplied 48 items to be assigned to 286 items of column 'units' (recycled leaving remainder of 46 items).

任何人都可以帮助我使用我的语法？

Can anyone help me with my syntax please?

PS：这是我的输入的输入，它看起来有点滑稽，虽然。

PS: Here's my dput for the input, it looks a little funny though.

> dput(c)
structure(list(`hh:mm` = c("11:00", "11:05", "11:10", "11:15", 
"11:20", "11:25", "11:30"), A = c(13.84, 12.54, 9.73, 9.2, 11.28, 
11.66, 10.38), V = c(470.16, 475.17, 476.2, 479.3, 482.22, 481.64, 
475.13), W = c(6509.88, 5959.22, 4632.14, 4410.89, 5437.78, 5614.56, 
4934), kA = c(14.89, 13.4, 10.36, 9.65, 12.03, 12.76, 11.99), 
    V = c(467.85, 474.52, 473.38, 482.79, 484.95, 479.95, 480.96
    ), kW = c(6964.38, 6358.89, 4905.86, 4659.67, 5835.33, 6124.56, 
    5760.44), A = c(15.74, 13.34, 10.38, 9.73, 12.24, 12.88, 
    11.5), kV = c(464.01, 473.13, 472.73, 479.09, 476.36, 476.86, 
    478.77), kW = c(7303.13, 6311.8, 4907.14, 4659.33, 5829.44, 
    6139.33, 5515.13)), .Names = c("hh:mm", "A", "V", "W", "kA", 
"V", "kW", "A", "kV", "kW"), row.names = c(NA, -7L), class = c("data.table", 
"data.frame"), .internal.selfref = <pointer: 0x00000000003b0788>)

推荐答案

你不是真的使用 data.table 正确。这里有一种方法可以使用：

You aren't really using data.table correctly. Here is one approach that can be used:

DT[, c("A2", "W2", "V2") := lapply(.SD, function(x) x*1000), .SDcols=c("kA", "kW", "kV")]
DT[, c("kA", "kW", "kV") := NULL]
DT
#    hh.mm     A      V       W    V.1   A.1    kW.1    A2      W2     V2
# 1: 11:00 13.84 470.16 6509.88 467.85 15.74 7303.13 14890 6964380 464010
# 2: 11:05 12.54 475.17 5959.22 474.52 13.34 6311.80 13400 6358890 473130
# 3: 11:10  9.73 476.20 4632.14 473.38 10.38 4907.14 10360 4905860 472730
# 4: 11:15  9.20 479.30 4410.89 482.79  9.73 4659.33  9650 4659670 479090
# 5: 11:20 11.28 482.22 5437.78 484.95 12.24 5829.44 12030 5835330 476360
# 6: 11:25 11.66 481.64 5614.56 479.95 12.88 6139.33 12760 6124560 476860
# 7: 11:30 10.38 475.13 4934.00 480.96 11.50 5515.13 11990 5760440 478770

注意：

• 您当前的解决方案已创建一个 data.table 名称。不好的做法。我错误地不在这里转换你的一个列，但我会离开这个练习更新你的答案。


• 如果你打算使用 data.table ，熟悉 set * 的一组函数。

• Your current solution has you creating a data.table with duplicated names. Bad practice. I made the mistake of not converting one of your columns here, but I'll leave that exercise of updating the answer to you.
• If you are going to use data.table, get familiar with the set* set of functions.

或者使用设置：

cols <- grep("^k", names(dt))
for (j in cols) {
    set(DT, i=NULL, j=j, value=DT[[j]]*1000)
}
# change names with `setnames` now.
setnames(DT, cols, gsub("^k", "", names(dt)[cols])

虽然我同意@Ananda不使用重复的名称。

Although I agree with @Ananda on not using duplicate names.

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