CODEWITHSUNDEEP
rdataframematch
Fabricio Gorleri
Fabricio Gorleri

Reputation: 23

Add the corresponding maximum value in a new column for each month

I've the following dataframe: 

df <- structure(list(locality_id = structure(c(3L, 1L, 2L, 4L), .Label = c("L2762894", 
"L3064193", "L6199859", "L8044094"), class = "factor"), month = structure(c(1L, 
3L, 1L, 2L), .Label = c("1", "2", "3"), class = "factor"), prec01 = c(105, 
109, 133, 79), prec02 = c(29, 34, 34, 35), prec03 = c(18, 48, 
42, 184)), class = "data.frame", row.names = c(NA, -4L))

Each locality has precipitation values for the first three months (prec01, prec02, prec03).

I would like to create an unique precipitation column matching the corresponding precipitation value of each month to the corresponding month (column month). Resulting in something as follows:

#   locality_id month prec
# 1    L6199859     1  105
# 2    L2762894     3  109
# 3    L3064193     1  133
# 4    L8044094     2  184

Any idea of how to do produce the above output?

Thank you!

Upvotes: 0

Views: 43

Answers (1)

Alessio
Alessio

Reputation: 920

If I understood your question you want to keep in the new prec column the corresponding value of precXX for that month?

In that case it seems to me that the example is not correct, but should instead be:

Starting dataframe:

  locality_id month prec01 prec02 prec03
1    L6199859     1    105     29     18
2    L2762894     3    109     34     48
3    L3064193     1    133     34     42
4    L8044094     2     79     35    184

Final dataframe:

 locality_id month prec
1    L6199859     1  105
2    L2762894     3   48
3    L3064193     1  133
4    L8044094     2   35

You may achieve that with matrix indexing:

i <- as.numeric(df$month)

df$prec <- df[,c(3,4,5)][cbind(seq_along(i), i)]

df[,c(-3,-4,-5)]

Upvotes: 1

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