英文:
Row-wise Boolean comparison of data
问题
我已经按适当的分组对数据进行了分组，我需要确保每个唯一的 Group1 和 Group2 组合下的 "x" 和 "y" 值相等。换句话说，我可以使用什么代码循环遍历这个数据集，并确保 A1x == A1y，A2x == A2y，等等。
以下是示例中的数据：
"Group1","Group2","group3","values""A" "1" x 10"A" "1" y 10"A" "2" x 15"A" "2" y 15
为了简化回答，以下是示例中的数据框：
d <- data.frame(Group1= c("A", "A", "A", "A"),Group2= c("1", "1", "2", "2"),group3= c("x", "y", "x", "y"),values= c(10, 10, 15, 15))
英文:
I have grouped my data by the appropriate grouping, and I need to be sure that "x" and "y" values equal each other for each unique combination of Group1 and Group2. In other words, what code could I use to cycle through this dataset and ensure that A1x == A1y and A2x == A2y, etc.
"Group1","Group2","group3","values""A" "1" x 10"A" "1" y 10"A" "2" x 15"A" "2" y 15
To help make the answer easier, here is the data.frame from the example
d <- data.frame(Group1= c("A", "A", "A", "A"),Group2= c("1", "1", "2", "2"),group3= c("x", "y", "x", "y"),values= c(10, 10, 15, 15))
答案1
得分: 4
使用dplyr，你可以做以下操作：
d %>%group_by(Group1, Group2) %>%mutate(cond = all(values == first(values)))
Group1 Group2 group3 values cond<fct> <fct> <fct> <dbl> <lgl>1 A 1 x 10 TRUE2 A 1 y 10 TRUE3 A 2 x 15 TRUE4 A 2 y 15 TRUE
或者：
d %>%group_by(Group1, Group2) %>%mutate(cond = n_distinct(values) == 1)
英文:
With dplyr, you can do:
d %>%group_by(Group1, Group2) %>%mutate(cond = all(values == first(values)))Group1 Group2 group3 values cond<fct> <fct> <fct> <dbl> <lgl>1 A 1 x 10 TRUE2 A 1 y 10 TRUE3 A 2 x 15 TRUE4 A 2 y 15 TRUE
Or:
d %>%group_by(Group1, Group2) %>%mutate(cond = n_distinct(values) == 1)
答案2
得分: 3
你也可以使用 pivot_wider 完成这个操作：
tidyr::pivot_wider(d, names_from='group3', values_from='values') %>%dplyr::mutate(eq=x==y)
英文:
You can also do this with pivot_wider:
tidyr::pivot_wider(d, names_from='group3', values_from='values') %>%dplyr::mutate(eq=x==y)
答案3
得分: 1
我认为你在将数据转换为长格式方面走得太远，也许这样更容易操作：
d %>%pivot_wider(names_from = group3, values_from = values) %>%mutate(is_equal = x == y)
英文:
I think you went too far into turning your data into a long format maybe this is easier to manipulate
d %>%pivot_wider(names_from = group3,values_from = values) %>%mutate(is_equal = x == y)
答案4
得分: 1
以下是使用基本的 R 解决方案，使用 ave() 来实现的部分：
d <- within(d, isequal <- as.logical(ave(values, Group1, Group2, FUN = function(v) v == unique(v))))
这样，数据框 d 中的 isequal 列将如下所示：
> dGroup1 Group2 group3 values isequal1 A 1 x 10 TRUE2 A 1 y 10 TRUE3 A 2 x 15 TRUE4 A 2 y 15 TRUE
请注意，这段代码使用 ave() 函数根据 Group1 和 Group2 列的组合来计算 values 列是否在组合内是唯一的，并将结果存储在 isequal 列中。
英文:
Here is a base R solution using ave() to make it
d <- within(d,isequal <- as.logical(ave(values,Group1,Group2,FUN = function(v) v==unique(v))))
such that
> dGroup1 Group2 group3 values isequal1 A 1 x 10 TRUE2 A 1 y 10 TRUE3 A 2 x 15 TRUE4 A 2 y 15 TRUE
答案5
得分: 0
另一种选项是，如果数据被正确分组并且每组有2行：
d$check <- rep(d$values[seq(1L, nrow(d), 2L)] == d$values[seq(2L, nrow(d), 2L)], each = 2L)
英文:
Another option if the data is grouped properly and has 2 rows for each group:
d$check <- rep(d$values[seq(1L,nrow(d),2L)]==d$values[seq(2L,nrow(d),2L)], each=2L)
答案6
得分: -1
一个简单的方法是合并具有组x和组y的子表格以比较数值。
> d[d$group3=="y",]# Group1 Group2 group3 values# 2 A 1 y 10# 4 A 2 y 15> merge(d[d$group3=="y",],d[d$group3=="x",],by=c("Group1","Group2"))# Group1 Group2 group3.x values.x group3.y values.y# 1 A 1 y 10 x 10# 2 A 2 y 15 x 15with(merge(d[d$group3=="y",], d[d$group3=="x",],by=c("Group1","Group2")),values.x==values.y)## [1] TRUE TRUE当然，你还有更高级的方法，但从简单开始并不是坏事。<details><summary>英文:</summary>A simple way would be to merge the sub tables with group x and group y to compare the values.> d[d$group3=="y",]# Group1 Group2 group3 values# 2 A 1 y 10# 4 A 2 y 15> merge(d[d$group3=="y",],d[d$group3=="x",],by=c("Group1","Group2"))# Group1 Group2 group3.x values.x group3.y values.y# 1 A 1 y 10 x 10# 2 A 2 y 15 x 15with(merge(d[d$group3=="y",], d[d$group3=="x",],by=c("Group1","Group2")),values.x==values.y)## [1] TRUE TRUEOf course you have fancier ways of doing it but it is not bad to start simple first</details>
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