向数据框中添加列 [英] Adding a column to a data.frame

问题描述

我有下面的data.frame。我想添加一个列，按照列1（ h_no ）对我的数据进行分类，这样h_no 1,2,3,4的第一个系列是class 1，第二个系列 h_no （1到7）是类2等，如最后一列所示。

  h_no h_freq h_freqsq 
 1 0.09091 0.008264628 1 
 2 0.00000 0.000000000 1 
 3 0.04545 0.002065702 1 
 4 0.00000 0.000000000 1 
 1 0.13636 0.018594050 2 
 2 0.00000 0.000000000 2 
 3 0.00000 0.000000000 2 
 4 0.04545 0.002065702 2 
 5 0.31818 0.101238512 2 
 6 0.00000 0.000000000 2 
 7 0.50000 0.250000000 2 
 1 0.13636 0.018594050 3 
 2 0.09091 0.008264628 3 
 3 0.40909 0.167354628 3 
 4 0.04545 0.002065702 3 
  

解决方案

您可以使用各种技术向列中添加列。下面的引号来自相关帮助文本的详细信息部分， [[。data.frame 。

数据框架可以以多种模式进行索引，当 [和 [[与单个向量索引 x [i] 或 x [[i]] ），它们将数据帧索引为列表。

  my.dataframe [new.col]<  -  a.vector 
 my.dataframe [ [new.col]]<  -  a.vector 
  

data.frame方法 $ ，将 x 作为列表

  my.dataframe $ new.col<  -  a.vector 
  

当 [和 [[与两个索引（ x [i，j] 和 x [[i，j]] ），它们像索引矩阵

  my.dataframe [，new.col]<  -  a.vector 
  

由于 data.frame 的方法假设如果您没有指定是否使用列或行，它将假定你的意思是列。

---

对于你的例子，这应该是有效的： p>

 ＃制作一些假数据
 your.df<  -  data.frame（no = c（1：4,1 ：7，1：5），h_freq = runif（16），h_freqsq = runif（16））
 
＃查找出现的地方，
来自<  -  which（your.df $没有== 1）
到<  -  c（（from-1）[ -  1]，nrow（your.df））＃到此为止，序列运行
 
＃generate一个序列（len）并根据其长度重复一个连续数len次
 get.seq<  -  mapply（from，to，1：length（from），FUN = function（x，y，z） {
 len<  -  length（seq（from = x [1]，to = y [1]））
 return（rep（z，times = len））
}）
 
＃当我们取消列表时，我们得到一个向量
 your.df $ group<  -  unlist（get.seq）
＃并将其附加到您的原始数据框架。因为这是
＃指定一个组，所以使它成为一个因素
 your.df $ group<  -  as.factor（your.df $ group）
 
 
否h_freq h_freqsq group 
 1 1 0.40998238 0.06463876 1 
 2 2 0.98086928 0.33093795 1 
 3 3 0.28908651 0.74077119 1 
 4 4 0.10476768 0.56784786 1 
 5 1 0.75478995 0.60479945 2 
 6 2 0.26974011 0.95231761 2 
 7 3 0.53676266 0.74370154 2 
 8 4 0.99784066 0.37499294 2 
 9 5 0.89771767 0.83467805 2 
 10 6 0.05363139 0.32066178 2 
 11 7 0.71741529 0.84572717 2 
 12 1 0.10654430 0.32917711 3 
 13 2 0.41971959 0.87155514 3 
 14 3 0.32432646 0.65789294 3 
 15 4 0.77896780 0.27599187 3 
 16 5 0.06100008 0.55399326 3 
  

I have the data.frame below. I want to add a column that classifies my data according to column 1 (h_no) in that way that the first series of h_no 1,2,3,4 is class 1, the second series of h_no (1 to 7) is class 2 etc. such as indicated in the last column.

h_no  h_freq  h_freqsq
1     0.09091 0.008264628 1
2     0.00000 0.000000000 1
3     0.04545 0.002065702 1
4     0.00000 0.000000000 1  
1     0.13636 0.018594050 2
2     0.00000 0.000000000 2
3     0.00000 0.000000000 2
4     0.04545 0.002065702 2
5     0.31818 0.101238512 2
6     0.00000 0.000000000 2
7     0.50000 0.250000000 2 
1     0.13636 0.018594050 3 
2     0.09091 0.008264628 3
3     0.40909 0.167354628 3
4     0.04545 0.002065702 3

解决方案

You can add a column to your data using various techniques. The quotes below come from the "Details" section of the relevant help text, [[.data.frame.

"Data frames can be indexed in several modes. When [ and [[ are used with a single vector index (x[i] or x[[i]]), they index the data frame as if it were a list."

my.dataframe["new.col"] <- a.vector
my.dataframe[["new.col"]] <- a.vector

"The data.frame method for $, treats x as a list"

my.dataframe$new.col <- a.vector

"When [ and [[ are used with two indices (x[i, j] and x[[i, j]]) they act like indexing a matrix"

my.dataframe[ , "new.col"] <- a.vector

Since the method for data.frame assumes that if you don't specify if you're working with columns or rows, it will assume you mean columns.

---

For your example, this should work:

# make some fake data
your.df <- data.frame(no = c(1:4, 1:7, 1:5), h_freq = runif(16), h_freqsq = runif(16))

# find where one appears and 
from <- which(your.df$no == 1)
to <- c((from-1)[-1], nrow(your.df)) # up to which point the sequence runs

# generate a sequence (len) and based on its length, repeat a consecutive number len times
get.seq <- mapply(from, to, 1:length(from), FUN = function(x, y, z) {
            len <- length(seq(from = x[1], to = y[1]))
            return(rep(z, times = len))
         })

# when we unlist, we get a vector
your.df$group <- unlist(get.seq)
# and append it to your original data.frame. since this is
# designating a group, it makes sense to make it a factor
your.df$group <- as.factor(your.df$group)


   no     h_freq   h_freqsq group
1   1 0.40998238 0.06463876     1
2   2 0.98086928 0.33093795     1
3   3 0.28908651 0.74077119     1
4   4 0.10476768 0.56784786     1
5   1 0.75478995 0.60479945     2
6   2 0.26974011 0.95231761     2
7   3 0.53676266 0.74370154     2
8   4 0.99784066 0.37499294     2
9   5 0.89771767 0.83467805     2
10  6 0.05363139 0.32066178     2
11  7 0.71741529 0.84572717     2
12  1 0.10654430 0.32917711     3
13  2 0.41971959 0.87155514     3
14  3 0.32432646 0.65789294     3
15  4 0.77896780 0.27599187     3
16  5 0.06100008 0.55399326     3

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