我需要从两个分类变量列中做一个频率表，其中一个是5年龄组，另一个是来自brfss2013数据集的健康状态（五个状态），从中我通过以下方式提取感兴趣的列：
> hlthgrpq1 <- brfss2013 %>% select(genhlth, X_ageg5yr)
从而生成一个两列框架，491775观察到2个变量。
genhlth X_ageg5yr
1 Fair Age 60 to 64
2 Good Age 50 to 54
3 Good Age 55 to 59
4 Very good Age 60 to 64
5 Good Age 65 to 69
我可以用'by'函数生成一个汇总表：
> by(hlthgrpq1$genhlth, hlthgrpq1$X_ageg5yr, summary)
hlthgrpq1$X_ageg5yr: Age 18 to 24
Excellent Very good Good Fair Poor NA's
6896 10266 7795 1873 303 69
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hlthgrpq1$X_ageg5yr: Age 25 to 29
Excellent Very good Good Fair Poor NA's
5779 8488 6521 1751 325 46
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hlthgrpq1$X_ageg5yr: Age 30 to 34
Excellent Very good Good Fair Poor NA's
6412 9958 7977 2295 496 75
----------------------------------------------------------------------------------------------------------------
hlthgrpq1$X_ageg5yr: Age 35 to 39
Excellent Very good Good Fair Poor NA's
6366 10169 8236 2637 638 61
----------------------------------------------------------------------------------------------------------------
hlthgrpq1$X_ageg5yr: Age 40 to 44
Excellent Very good Good Fair Poor NA's
6689 11130 9193 3334 1067 95
----------------------------------------------------------------------------------------------------------------
hlthgrpq1$X_ageg5yr: Age 45 to 49
Excellent Very good Good Fair Poor NA's
7051 12278 10611 4343 1815 112
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hlthgrpq1$X_ageg5yr: Age 50 to 54
Excellent Very good Good Fair Poor NA's
8545 15254 13761 6354 3120 139
----------------------------------------------------------------------------------------------------------------
hlthgrpq1$X_ageg5yr: Age 55 to 59
Excellent Very good Good Fair Poor NA's
8500 16759 15394 7643 3998 197
----------------------------------------------------------------------------------------------------------------
hlthgrpq1$X_ageg5yr: Age 60 to 64
Excellent Very good Good Fair Poor NA's
8283 16825 16266 8101 3955 229
----------------------------------------------------------------------------------------------------------------
hlthgrpq1$X_ageg5yr: Age 65 to 69
Excellent Very good Good Fair Poor NA's
7479 15764 15600 7749 3200 205
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hlthgrpq1$X_ageg5yr: Age 70 to 74
Excellent Very good Good Fair Poor NA's
5491 11943 13125 6491 2721 196
----------------------------------------------------------------------------------------------------------------
hlthgrpq1$X_ageg5yr: Age 75 to 79
Excellent Very good Good Fair Poor NA's
3320 8501 10128 5545 2426 173
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hlthgrpq1$X_ageg5yr: Age 80 or older
Excellent Very good Good Fair Poor NA's
3697 10285 14400 8116 3695 322
这就是我被卡住的地方。我已经尝试了几个小时试图到达这里：
Results obtained via spreadsheet.
谢谢你的帮助。
（这是针对特定的任务，所以我只能使用dplyr和ggplot2，所以，没有reshape2或者tidyr。）
首先：对于将来的发布，最好始终包含样本数据。请参阅here how to provide a minimal reproducible example/attempt，包括样本数据。
基础R的解决方案
as.data.frame.matrix(t(table(df)));
# Fair Good Very good
#Age 50 to 54 0 1 0
#Age 55 to 59 0 1 0
#Age 60 to 64 1 0 1
#Age 65 to 69 0 1 0
或者类似这样的tidyverse方法？
library(tidyverse);
df %>% count(genhlth, X_ageg5yr) %>% spread(genhlth, n);
## A tibble: 4 x 4
# X_ageg5yr Fair Good `Very good`
# <fct> <int> <int> <int>
#1 Age 50 to 54 NA 1 NA
#2 Age 55 to 59 NA 1 NA
#3 Age 60 to 64 1 NA 1
#4 Age 65 to 69 NA 1 NA
或者，如果你坚持只使用dplyr而不是tidyr，你可以这样做：
df2 <- df %>%
count(genhlth, X_ageg5yr);
df2 <- as.data.frame.matrix(xtabs(n ~ X_ageg5yr + genhlth, data = df2));
# Fair Good Very good
#Age 50 to 54 0 1 0
#Age 55 to 59 0 1 0
#Age 60 to 64 1 0 1
#Age 65 to 69 0 1 0
这基本上归结为一个从长到长的重新格式，SO围绕该主题进行了大量讨论（例如here）。
df <- read.table(text =
"genhlth X_ageg5yr
Fair 'Age 60 to 64'
Good 'Age 50 to 54'
Good 'Age 55 to 59'
'Very good' 'Age 60 to 64'
Good 'Age 65 to 69'", header = T)