为每个连续序列创建一个组号

本文介绍了为每个连续序列创建一个组号的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我在下面有data.frame.我想添加一列"g"，该列根据 h_no 列中的连续序列对我的数据进行分类.也就是说，h_no 1、2、3、4 的第一个序列是组1， h_no 的第二个序列(1到7)是组2，依此类推.，如最后一列"g"所示.

I have the data.frame below. I want to add a column 'g' that classifies my data according to consecutive sequences in column h_no. That is, the first sequence of h_no 1, 2, 3, 4 is group 1, the second series of h_no (1 to 7) is group 2, and so on, as indicated in the last column 'g'.

h_no h_freq h_freqsq g 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

推荐答案

您可以使用多种技术在数据中添加一列.下面的引号来自相关帮助文本 [[.. data.frame .

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.

可以在几种模式下对数据帧建立索引.当 [和 [[]与单个矢量索引( x [i] 或 x [[i]] )，它们将数据框索引为列表.

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

$ 的data.frame方法将 x 视为列表

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

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

当 [和 [[]与两个索引( x [i，j] 和 x [[i，j]] )，它们就像索引矩阵一样

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

由于 data.frame 的方法假定如果您未指定要使用列还是行，则将假定您是指列.

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.

以您的示例为例，这应该可行:

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