传递带有名称的向量以进行变异以创建多个新列 [英] Pass a vector with names to mutate to create multiple new columns
问题描述
我正尝试使用包含正确答案的向量重新编码答案。我做了一个for循环，使用带有新列可能名称的向量在每个循环处创建一个新列（带有编码答案）。
I'm trying to recode answers using a vector that contains the correct answers. I made a for loop that create a new column (with the coded answer) at each loop using a vector with the possible names for the new columns.
但是，似乎突变不会接收带有名称的向量。我尝试了一些不同的向量和某些paste0（）组合，但似乎没有任何效果。
However, it seems that mutate does not receive vectors with names. I've tried some different vectors and some paste0() combinations but nothing seem to work.
这是我的可复制代码：
library(dplyr)
library(tibble)
correct = c(4, 5, 2, 2, 2, 3, 3, 5, 4, 5, 2, 1, 3, 4, 2, 2, 2, 4, 3, 1, 1, 5, 4, 1, 3, 2)
sub1 = c(3, 5, 1, 5, 4, 3, 2, 5, 4, 3, 4, 4, 4, 1, 5, 1, 4, 3, 3, 4, 3, 2, 4, 2, 3, 4)
df = t(data.frame(sub1))
colnames(df) = paste0("P", 1:26)
new_names = paste0("P", 1:26, "_coded")
for(i in 1:26){
df = as.tibble(df) %>%
mutate(new_names = case_when(.[i] == correct[i] ~ 1,
.[i] != correct[i] ~ 0,
T ~ 9999999))
print(df) # to know what's going on.
}
此外，我知道.dot可以在向量（我认为），但在mutate（）中时，我不太了解如何与case_一起使用。
Also, I know that .dots can receive names in a vector (I think), but I don't quite understand how to use it with case_when inside mutate().
其他使用重新编码的值创建新列的方法是也欢迎
Others ways to create new columns with the recoded value are welcome also
更新：
我的预期输出将是带有26个新列的原始数据框，P1_COD：P26_COD并带有可能的值1（如果正确）和0（如果不正确）。
UPDATE: My expected output would be the original data frame with 26 new columns, P1_COD:P26_COD with possible values 1 (if correct) and 0 (if incorrect).
像这样的事情（我刚刚创建了四个带有1和0的列）。
Something like this (I just created four columns with 1s and 0s as an example).
df %>%
mutate(P1_COD = 1,
P2_COD = 0,
P3_COD = 1,
P4_COD = 1)
推荐答案
格式不是 dplyr 将能最好地处理的格式。我建议将数据重组为纵向格式，然后case_when变得琐碎且不需要for循环。
The data is not in a format that dplyr will handle best. I would suggest restructuring your data to longitudinal format, and then the case_when becomes trivial and no for loop is required.
有关提迪尔的其他文档，请参见 = http://tidyr.tidyverse.org/articles/tidy-data.html rel = nofollow noreferrer> tidyverse.org文档
see other documentation for tidyr regarding data format at tidyverse.org documentation
这是包括您的样本数据的纵向格式的示例。我还添加了其他一些具有随机答案的主题。
Here is an example of the "longitudinal" format including your sample data. I also added a couple of other subjects with random answers.
library(tidyverse)
responses <- data_frame(
subject = rep(1:3, each = 26),
qNum = rep(1:26, 3),
response = c(sub1,
sample(5, 26, replace = T),
sample(5, 26, replace = T)))
可以创建然后合并答案：
The answers can be created and then merged:
answers <- data_frame(
qNum = 1:26,
answer = correct)
df <- left_join(responses, answers)
接下来，使用 dplyr :: case_when ：
df <- df %>% mutate(score = case_when(response == answer ~ 1,
TRUE ~ 0))
注意： TRUE〜0 最初可能会造成混淆。如果第一个条件为FALSE，它将告诉如何处理剩余的值。结果df / tibble：
note: the TRUE ~ 0 may be confusing at first. It tells what to do with the remaining values, if the first condition is FALSE. The resulting df/tibble:
# A tibble: 26 x 5
subject qNum response answer score
<dbl> <int> <dbl> <dbl> <dbl>
1 1 1 3 4 0
2 1 2 5 5 1
3 1 3 1 2 0
4 1 4 5 2 0
5 1 5 4 2 0
6 1 6 3 3 1
7 1 7 2 3 0
8 1 8 5 5 1
9 1 9 4 4 1
10 1 10 3 5 0
# ... with 16 more rows
如果要将其转换为宽格式，请使用 tidyr :: spread ：
If you want to convert this to "wide" format, use tidyr::spread:
df %>%
select(-response, -answer) %>%
spread(qNum, score, sep = ".")
# A tibble: 3 x 27
subject qNum.1 qNum.2 qNum.3 qNum.4 qNum.5 qNum.6 qNum.7 qNum.8 qNum.9 qNum.10
* <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 0 1 0 0 0 1 0 1 1 0
2 2 0 0 0 0 1 0 0 0 0 0
3 3 0 0 0 0 1 0 0 0 0 0
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