英文:
R - How to put aggregated row results as columns
问题
假设有一个名为df的数据框：
age category12 A15 B12 A13 C14 B14 D
你想使用聚合函数来找出每个年龄（age）中每个类别（category）{A, B, C, D}的出现次数。A、B、C和D的计数分别作为列，因此输出的数据框应如下所示：
age A B C D12 2 0 0 013 0 0 1 014 0 1 0 115 0 1 0 0
你尝试的代码如下：
agdf <- aggregate(df, by=list(df$age, df$category), FUN=length)
但这只会得到以下结果：
age category x12 A 215 B 114 B 113 C 114 D 1
另一个问题是原始的df数据框还有其他列，但为简化起见，它们已被省略。然而，使用FUN=length的聚合方法将所有这些其他列都变成与"x"列相同的计数值。你如何保留这些值？
例如：
age category x y z12 A 2 2 215 B 1 1 114 B 1 1 113 C 1 1 114 D 1 1 1
但我只想要y和z保持它们的原始值，只需要一个计数列x。如何将数据框结构调整为所需的形式？
解决方法：
你可以使用reshape2包中的dcast函数来实现你想要的结果。首先，确保你已经加载了reshape2包。然后，可以按照以下方式操作：
library(reshape2)# 使用dcast函数result <- dcast(df, age ~ category, value.var = "category", fun.aggregate = length, fill = 0)# 如果需要保留其他列，可以使用merge函数# 假设df包括其他列y和zdf <- merge(df, result, by = "age", all = TRUE)
这将生成你所需的数据框，其中"A"、"B"、"C"和"D"的计数作为列，而其他列（例如y和z）也被保留。
英文:
Suppose there is dataframe df
age category12 A15 B12 A13 C14 B14 D
I want to use aggregate to find the number of occurrences in each category {A, B, C, D} for each age. The number of A, B, C and D respectively are established as columns so the output data frame should look like
age A B C D12 2 0 0 013 0 0 1 014 0 1 0 115 0 1 0 0
Attempt
agdf <- aggregate(df, by=list(df$age, df$category), FUN=length)
But doing this only gives me
age category x12 A 215 B 114 B 113 C 114 D 1
Another problem is that the original df has other columns but they have been omitted for simplicity. Yet with this aggregate approach using FUN=length, it turns all those other columns into the same count value as x. How can I keep those values?
E.g.
age category x y z12 A 2 2 215 B 1 1 114 B 1 1 113 C 1 1 114 D 1 1 1
but I want y and z to keep their original values, only need 1 count column x
How to massage it to the desired structure?
答案1
得分: 1
xtabs(~., df1)
categoryage A B C D12 2 0 0 013 0 0 1 014 0 1 0 115 0 1 0 0
table(df1)
category
age A B C D
12 2 0 0 0
13 0 0 1 0
14 0 1 0 1
15 0 1 0 0
reshape2::dcast(df1, age~category)
age A B C D
1 12 2 0 0 0
2 13 0 0 1 0
3 14 0 1 0 1
4 15 0 1 0 0
pivot_wider(df1, id_cols = age, names_from = category,
values_from = category, values_fn = length, values_fill = 0)
A tibble: 4 × 5
age A B C D
1 12 2 0 0 0
2 15 0 1 0 0
3 13 0 0 1 0
4 14 0 1 0 1
英文:
xtabs(~., df1)categoryage A B C D12 2 0 0 013 0 0 1 014 0 1 0 115 0 1 0 0table(df1)categoryage A B C D12 2 0 0 013 0 0 1 014 0 1 0 115 0 1 0 0reshape2::dcast(df1, age~category)age A B C D1 12 2 0 0 02 13 0 0 1 03 14 0 1 0 14 15 0 1 0 0pivot_wider(df1, id_cols = age, names_from = category,values_from = category, values_fn = length, values_fill = 0)# A tibble: 4 × 5age A B C D<int> <int> <int> <int> <int>1 12 2 0 0 02 15 0 1 0 03 13 0 0 1 04 14 0 1 0 1
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