This question already has an answer here:
这个问题在这里已有答案：
- Finding the column number and value the of second highest value in a row 1 answer
- 查找列号并将值排在第1行的第二个最高值
Sorry if I am asking a trivial question, but the fact is I have spent a few hours reading answers in this data base and could not find what I am looking for.
对不起，如果我问一个微不足道的问题，但事实是我花了几个小时阅读这个数据库中的答案，但找不到我要找的东西。
I have a dataframe similar to this
我有一个与此类似的数据帧
df=data.frame(v1=c(24,15, 0, 7,36,10), c1=c(22,15,0,0,28,11), v2=c(0,10,0,19,0,0), c2=c(0,7,0,22,0,0), v3=c(54,22,28,55,62,38), c3=c(44,23,22,66,71,44))
(The original, of course, has many more rows and columns)
（当然，原来有更多的行和列）
I would like to create two columns with the maximum and the second highest values of all the "v" columns.
我想创建两列，其中包含所有“v”列的最大值和第二个最高值。
For the maximum, this works:
最大化，这有效：
df$max.v=mapply(FUN=max, df$v1, df$v2, df$v3, na.rm=TRUE)
But I cannot find a way to do it for the second highest value. It probably needs some kind of function, but I could not find how to do it.
但我无法找到第二高价值的方法。它可能需要某种功能，但我找不到如何做到这一点。
1 个解决方案
#1
1
Note that the accepted answer in the question linked by @krlmlr is dubious, because apply can break data frames. It doesn't matter so much in this case, because all the columns must be numeric for the question to make sense, but I prefer to err on the safe side.
请注意，@ krlmlr链接的问题中接受的答案是可疑的，因为apply可能会破坏数据帧。在这种情况下，这并不重要，因为所有列都必须是数字才能使问题有意义，但我宁愿在安全方面犯错。
Instead, use do.call with mapply, and persuade it to treat a df as a list:
相反，使用do.call与mapply，并说服它将df视为列表：
do.call(mapply, c(function(...) sort(c(...), dec=TRUE)[1:2],
df[grepl("v", names(df))]))
#1
1
Note that the accepted answer in the question linked by @krlmlr is dubious, because apply can break data frames. It doesn't matter so much in this case, because all the columns must be numeric for the question to make sense, but I prefer to err on the safe side.
请注意，@ krlmlr链接的问题中接受的答案是可疑的，因为apply可能会破坏数据帧。在这种情况下，这并不重要，因为所有列都必须是数字才能使问题有意义，但我宁愿在安全方面犯错。
Instead, use do.call with mapply, and persuade it to treat a df as a list:
相反，使用do.call与mapply，并说服它将df视为列表：
do.call(mapply, c(function(...) sort(c(...), dec=TRUE)[1:2],
df[grepl("v", names(df))]))