Question

是否可以仅为分裂变量的某些值返回ddply结果？例如，使用数据框example：

example <- structure(list(shape = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L, 2L, 3L, 3L, 3L, 3L, 3L), .Label = c("circle", "square", "triangle"
), class = "factor"), property = structure(c(1L, 3L, 2L, 1L, 
2L, 3L, 1L, 1L, 1L, 1L, 2L, 3L, 1L, 1L), .Label = c("color", 
"intensity", "size"), class = "factor"), value = structure(c(5L, 
2L, 1L, 5L, 4L, 1L, 5L, 6L, 6L, 7L, 4L, 3L, 6L, 5L), .Label = c("3", 
"5", "6", "7", "blue", "green", "red"), class = "factor")), .Names = c("shape", 
"property", "value"), class = "data.frame", row.names = c(NA, 
-14L))

看起来像这样

    shape     property  value
1   circle    color     blue
2   circle    size      5
3   circle    intensity 3
4   circle    color     blue
5   square    intensity 7
6   square    size      3
7   square    color     blue
8   square    color     green
9   square    color     green
10  triangle  color     red
11  triangle  intensity 7
12  triangle  size      6
13  triangle  color     green
14  triangle  color     blue

我想返回一个数据框，其中包含每种具有特定颜色的形状的数量，如下所示：

    shape    property  blue green   red
1   circle   color     2    0       0
2   square   color     1    2       0
3   triangle color     1    1       1

然而，我似乎无法让这个正确归还！我已经使用了类似的东西：

ColorSummary <- ddply(example,.(shape,property="color"), function(example) summary(example$value))

但是这会返回一个数据框，其中包含所有其他唯一value的列（来自属性size和intensity，我不想要）：

    shape     property      3   5   6   7   blue    green   red
1   circle    color         1   1   0   0   2       0       0
2   square    NA            1   0   0   1   1       2       0
3   triangle  NA            0   0   1   1   1       1       1

我做错了什么 - 是否有办法返回数据框，就像我展示的第一个结果一样？

此外，虽然这是一个小而快的例子，但我的“真实”数据要大得多，需要很长时间才能计算出来。仅通过限制property="color"来改善ddply的速度吗？

编辑：感谢您的回答！不幸的是，我过度简化了情况，我不确定dcast解决方案是否对我有用。让我解释一下 - 我实际上正在使用数据框example2：

example2 <- structure(list(factory = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L), .Label = c("A", 
"B"), class = "factor"), shape = structure(c(1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L), .Label = c("circle", 
"square", "triangle"), class = "factor"), property = structure(c(1L, 
3L, 2L, 1L, 2L, 3L, 1L, 1L, 1L, 1L, 2L, 3L, 1L, 1L, 1L, 3L, 2L
), .Label = c("color", "intensity", "size"), class = "factor"), 
    value = structure(c(5L, 2L, 1L, 5L, 4L, 1L, 5L, 6L, 6L, 7L, 
    4L, 3L, 6L, 5L, 5L, 2L, 1L), .Label = c("3", "5", "6", "7", 
    "blue", "green", "red"), class = "factor")), .Names = c("factory", 
"shape", "property", "value"), class = "data.frame", row.names = c(NA, 
-17L))

我试图按factory和shape进行拆分。我使用ddply：

的解决方案很混乱

ColorSummary2 <- ddply(example2,.(factory,shape,property="color"), function(example2) summary(example2$value))

给出了

    factory shape   property    3   5   6   7   blue    green   red
1   A   circle  color   1   1   0   0   2   0   0
2   A   square  NA  1   0   0   1   1   2   0
3   A   triangle    NA  0   0   1   1   1   1   1
4   B   circle  NA  1   1   0   0   1   0   0

但是我想要返回的是（对于凌乱的表格，我在这里格式化表格时遇到了麻烦）：

    factory shape   property        blue    green   red
1   A   circle      color           2       0       0
2   A   square      NA              1       2       0
3   A   triangle    NA              1       1       1   
4   B   circle      NA              1       0       0

这可能吗？

编辑2：对于所有的编辑都抱歉，我过分简化了我的情况。这是一个更复杂的数据框，更接近我的真实例子。这个列有一个state列，我不想用它来进行拆分。我可以用ddply做这个（乱七八糟），但是我可以使用dcast忽略state吗？

example3 <- structure(list(state = structure(c(1L, 2L, 1L, 2L, 1L, 2L, 1L, 
2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L), .Label = c("CA", "FL"
), class = "factor"), factory = structure(c(1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L), .Label = c("A", 
"B"), class = "factor"), shape = structure(c(1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L), .Label = c("circle", 
"square", "triangle"), class = "factor"), property = structure(c(1L, 
3L, 2L, 1L, 2L, 3L, 1L, 1L, 1L, 1L, 2L, 3L, 1L, 1L, 1L, 3L, 2L
), .Label = c("color", "intensity", "size"), class = "factor"), 
    value = structure(c(5L, 2L, 1L, 5L, 4L, 1L, 5L, 6L, 6L, 7L, 
    4L, 3L, 6L, 5L, 5L, 2L, 1L), .Label = c("3", "5", "6", "7", 
    "blue", "green", "red"), class = "factor")), .Names = c("state", 
"factory", "shape", "property", "value"), class = "data.frame", row.names = c(NA, 
-17L))

Answer 1

使用dcast中的reshape2：

dcast(...~value,data=subset(example,property=='color'))
Aggregation function missing: defaulting to length
     shape property blue green red
1   circle    color    2     0   0
2   square    color    1     2   0
3 triangle    color    1     1   1

修改

使用第二个数据集示例：

dcast(...~value,data=subset(example2,property=='color'))
Aggregation function missing: defaulting to length
  factory    shape property blue green red
1       A   circle    color    2     0   0
2       A   square    color    1     2   0
3       A triangle    color    1     1   1
4       B   circle    color    1     0   0

ddply只有一定的分裂变量值

1 个答案:

修改