R-根据与现有列匹配的值索引创建新的列数据框 [英] R - create new column dataframe based on index of value match with existing column

问题描述

我有一个名为imp2的数据帧(大约6,000行)，其中有9列标记为'savres1'...'savres9'的列.这些列中每列的值为1或0.对于每一行，这些列中只有1的值为1(其余均为0).

I have a dataframe called imp2 (with about 6,000 rows) for which there are 9 columns labeled 'savres1'...'savres9'. Values in each of these columns are either 1 or 0. For each row, the value of only 1 of these columns is 1 (rest are 0).

我想创建一个名为'savres'的新列，在该列中放置一个1到9之间的值，该值与保存1值的列的索引(在这组9列中)匹配.例如.如果'savres7'为1，其余列为0，则savres应该等于7.

I'd like to create a new column called 'savres' into which I would put a value between 1 and 9 that matches the index of the column (within this set of 9 columns) which holds the 1 value. E.g. if 'savres7' is 1 and the rest of these columns are 0, then savres should equal 7.

我使用了which，match和%in%的几种变体来找到值，但没有得到我期望的输出.

I've used several variations of which, match and %in% to find the value, but I'm not getting the output I expect.

下面是最近的试用功能，以及在数据框中创建的示例行集.

Below is a recent trial function and a sample set of rows that are created in the dataframe.

最近的尝试解决方案:

imp2 <- within(imp2, savres <- which(c(savres1, savres2, savres3, savres4, savres5, savres6, savres7, savres8, savres9) == 1) %% 9 + 1)

结果:

(注意，我没有包括所有列，但是从这里的第4列中，您可以看到该函数不起作用-第4行中的'savres'值应为3，而第1行和第2行的值应为不是4或2)

(Note, I didn't include all columns, but from the 4 here, you can see that the function isn't working - the 'savres' value in row 4 should be 3 and those for rows 1 and 2 should not be 4 or 2)

sample <- head(imp2[c('savres1','savres2','savres3','savres4')],4)

savres | savres1 | savres2 | savres3 | savres4
4      |     0   |    0    |    0    |    0
2      |   0     |    0    |    0    |    0
1      |   1     |    0    |    0    |   0
1      |   0     |   0     |  1      |   0

任何人都可以推荐解决方案吗?

Can anyone recommend a solution?

我认为这应该是一个相当简单的操作，但是我对尝试过的许多方法感到沮丧.

I thought this should be a fairly simple operation, but I've been frustrated with a number of approaches I've tried.

推荐答案

我们可以使用max.col

i1 <- grep("^savre", names(df1))
transform(df1, savre = (max.col(df1[i1], "first"))* !!rowSums(df1[i1]))
#    col1 col2 savres1 savres2 savres3 savres4 savre
#1    1    2       0       0       0       0     0
#2    2    3       0       0       0       0     0
#3    3    4       1       0       0       0     1
#4    4    5       0       0       1       0     3

数据

df1 <- data.frame(col1 = 1:4, col2 = 2:5, savres1 = c(0, 0, 1,0), 
   savres2 = 0, savres3 = c(0, 0, 0, 1), savres4 = 0)

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