问题描述
我继承了一个以不寻常方式编码的数据集.我想学习一种不那么冗长的重塑方式.数据框如下所示:
I inherited a data set coded in an unusual way. I would like to learn a less verbose way of reshaping it. The data frame looks like this:
# Input.
participant = c(rep("John",6), rep("Mary",6))
day = c(rep(1,3), rep(2,3), rep(1,3), rep(2,3))
likes = c("apples", "apples", "18", "apples", "apples", "7", "bananas", "bananas", "24", "bananas", "bananas", "3")
question = rep(c(1,1,0),4)
number = c(rep(18,3), rep(7,3), rep(24,3), rep(3,3))
df = data.frame(participant, day, question, likes)
participant day question likes
1 John 1 1 apples
2 John 1 1 apples
3 John 1 0 18
4 John 2 1 apples
5 John 2 1 apples
6 John 2 0 7
7 Mary 1 1 bananas
8 Mary 1 1 bananas
9 Mary 1 0 24
10 Mary 2 1 bananas
11 Mary 2 1 bananas
12 Mary 2 0 3
如您所见，likes 列是异构的.当问题等于0时，喜欢传达的是参与者选择的数字，而不是他们喜欢的水果.所以我想在一个新列中重新编码如下:
As you can see, the column likes is heterogeneous. When question equals 0, likes conveys a number chosen by the participants, not their preferred fruit. So I would like to re-code it in a new column as follows:
participant day question likes number
1 John 1 1 apples 18
2 John 1 1 apples 18
3 John 1 0 18 18
4 John 2 1 apples 7
5 John 2 1 apples 7
6 John 2 0 7 7
7 Mary 1 1 bananas 24
8 Mary 1 1 bananas 24
9 Mary 1 0 24 24
10 Mary 2 1 bananas 3
11 Mary 2 1 bananas 3
12 Mary 2 0 3 3
我当前使用基本 R 的解决方案涉及对初始数据框进行子集化、创建查找表、更改列名称，然后将查找表与原始数据框合并.但这涉及几个步骤，我担心应该有一个更简单的解决方案.我认为 tidyr 可能是答案，但我不知道如何使用它在一列(喜欢)条件其他列(天 和 问题).
My current solution with base R involves subsetting the initial data frame, creating a lookup table, changing the column names and then merging the lookup table with the original data frame. But this involves several steps and I worry that there should be a simpler solution. I think that tidyr might be the answer, but I don't know how to use it to spread values in one column (likes) conditional other columns (day and question).
你有什么建议吗?非常感谢！
Do you have any suggestions? Thanks a lot!
推荐答案
使用上面的数据集，可以尝试以下操作.您按 participant 和 day 对数据进行分组，并为每个组查找具有 question == 0 的行.
Using the data set above, you can try the following. You group your data by participant and day and look for a row with question == 0 for each group.
library(dplyr)
group_by(df, participant, day) %>%
mutate(age = as.numeric(as.character(likes[which(question == 0)])))
或者按照 alistaire 的建议，您也可以使用 grep().
Or as alistaire suggested, you can use grep() too.
group_by(df, participant, day) %>%
mutate(age = as.numeric(grep('\\d+', likes, value = TRUE)))
# participant day question likes age
# (fctr) (dbl) (dbl) (fctr) (dbl)
#1 John 1 1 apples 18
#2 John 1 1 apples 18
#3 John 1 0 18 18
#4 John 2 1 apples 7
#5 John 2 1 apples 7
#6 John 2 0 7 7
#7 Mary 1 1 bananas 24
#8 Mary 1 1 bananas 24
#9 Mary 1 0 24 24
#10 Mary 2 1 bananas 3
#11 Mary 2 1 bananas 3
#12 Mary 2 0 3 3
如果你想使用data.table，你可以这样做:
If you want to use data.table, you can do:
library(data.table)
setDT(df)[, age := as.numeric(as.character(likes[which(question == 0)])),
by = list(participant, day)]
注意
目前的数据集是一个新的.Jota 的回答适用于已删除的数据集.
The present data set is a new one. Jota's answer works for the deleted data set.
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