行和列中有多个项目的交叉表

本文介绍了行和列中有多个项目的交叉表的处理方法，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

假设我的数据来自于对名为 metric_1 的变量按四个不同组(xi、delta、scenario 和 model) 给出一个名为 mean_metric1 的新变量.数据如下所示.

Let's say I have data that comes from averaging a variable called metric_1 by four different groups (xi, delta, scenario and model) giving a new variable called mean_metric1. The data would look like the following.

set.seed(777) xi = c(2, 4 ) delta = c( 0.5, 1 ) scenario = c("scenario_1","scenario_2") model <- c("model_1", "model_2", "model_3") df <- expand.grid(xi = xi, delta = delta, scenario = scenario, model = model) df$mean_metric1 <- rnorm(nrow(df), 100 , 25) xi delta scenario model mean_metric1 1 2 0.5 scenario_1 model_1 112.24466 2 4 0.5 scenario_1 model_1 90.03647 3 2 1.0 scenario_1 model_1 112.77091 4 4 1.0 scenario_1 model_1 90.02970 5 2 0.5 scenario_2 model_1 140.96715 6 4 0.5 scenario_2 model_1 115.53185 7 2 1.0 scenario_2 model_1 105.06761 8 4 1.0 scenario_2 model_1 127.72344 9 2 0.5 scenario_1 model_2 94.84438 10 4 0.5 scenario_1 model_2 90.52587 11 2 1.0 scenario_1 model_2 92.39346 12 4 1.0 scenario_1 model_2 101.35406 13 2 0.5 scenario_2 model_2 52.97673 14 4 0.5 scenario_2 model_2 99.15609 15 2 1.0 scenario_2 model_2 157.78737 16 4 1.0 scenario_2 model_2 124.30850 17 2 0.5 scenario_1 model_3 124.11520 18 4 0.5 scenario_1 model_3 86.39669 19 2 1.0 scenario_1 model_3 116.78058 20 4 1.0 scenario_1 model_3 112.52047 21 2 0.5 scenario_2 model_3 49.23410 22 4 0.5 scenario_2 model_3 105.69380 23 2 1.0 scenario_2 model_3 80.42437 24 4 1.0 scenario_2 model_3 131.83980

我想使用列中的两个变量和行中的两个变量根据这些信息创建一个交叉表，如对象 wanted 所示.

I want to create a cross table out of this information using two variables in the columns and two variables in the rows as illustrated in object wanted.

wanted <- structure(list(V1 = c("delta_0.5", "xi_2", "112.24466", "94.84438","124.1152", "140.96715", "52.97673", "49.2341"), V2 = c("delta_0.5","xi_4", "90.03647", "90.52587", "86.39669", "115.53185", "99.15609","105.6938"), V3 = c("delta_1", "xi_2", "112.77091", "92.39346","116.78058", "105.06761", "157.78737", "80.42437"), V4 = c("delta_1","xi_4", "90.0297", "101.35406", "112.52047", "127.72344", "124.3085","131.8398")), class = "data.frame", row.names = c(" ", "", "model_1_scenario_1","model_2_scenario_1", "model_3_scenario_1", "model_1_scenario_2","model_2_scenario_2", "model_3_scenario_2")) V1 V2 V3 V4 delta_0.5 delta_0.5 delta_1 delta_1 xi_2 xi_4 xi_2 xi_4 model_1_scenario_1 112.24466 90.03647 112.77091 90.0297 model_2_scenario_1 94.84438 90.52587 92.39346 101.35406 model_3_scenario_1 124.1152 86.39669 116.78058 112.52047 model_1_scenario_2 140.96715 115.53185 105.06761 127.72344 model_2_scenario_2 52.97673 99.15609 157.78737 124.3085 model_3_scenario_2 49.2341 105.6938 80.42437 131.8398

推荐答案

您可以将 ftable 与 xtabs 结合使用:

You can use ftable in conjunction with xtabs:

ftable(xtabs(mean_metric1 ~ scenario + model + xi + delta, df), col.vars=3:4) xi 2 4 delta 0.5 1 0.5 1 scenario model scenario_1 model_1 112.24466 112.77091 90.03647 90.02970 model_2 94.84438 92.39346 90.52587 101.35406 model_3 124.11520 116.78058 86.39669 112.52047 scenario_2 model_1 140.96715 105.06761 115.53185 127.72344 model_2 52.97673 157.78737 99.15609 124.30850 model_3 49.23410 80.42437 105.69380 131.83980

xtabs 是必需的，因为直接使用 ftable 只会给出组合的计数.请注意，xtabs 将 sum 重复组合.

xtabs is required since using ftable directly would just give a count of the the combinations. Note that xtabs will sum duplicated combinations.