如何申请 3 function 新建 dataframe

Question

I'm new here on StackOverflow. I would like to apply 3 function to a dataframe in order to create a new dataframe.

emiscore$rank19<-rank(-emiscore$"2019")
emi_P_19<-filter(emiscore,rank19<31)
emi_P_19<-emi_P_19[order(emi_P_19$Name),]

The top 10 lines of emi_P_19 appears as following:

structure(list(Name = c("LA Z BOY", "1 800 FLOWERS.COM 'A'", 
"AGEAS (EX FORTIS)", "AGFA GEVAERT", "AIR FRANCE KLM", "ANHEUSER BUSCH INBEV"
), DATATYPE = c("TRESGENERS", "TRESGENERS", "TRESGENERS", "TRESGENERS", 
"TRESGENERS", "TRESGENERS"), `2019` = c(0, 0, NA, NA, NA, NA), 
    `2018` = c(8.33, 0, 22.15, 64.46, 97.92, 58.47), `2017` = c(0, 
    0, 0, 63.11, 97.83, 49.14), `2016` = c(0, 0, 0, 58.65, 95.83, 
    61.46), `2015` = c(NA, NA, 0, 64.89, 93.27, 67.71), `2014` = c(NA, 
    NA, 0, 60.26, 94.57, 59.78), `2013` = c(NA, NA, 0, 64.63, 
    96.74, 77.17), `2012` = c(NA, NA, 0, 67.86, 98.96, 75), `2011` = c(NA, 
    NA, 0, 67.07, 96.81, 70.93), `2010` = c(NA, NA, 17.05, 71.25, 
    98.98, 88.46), `2009` = c(NA, NA, 11.59, 68.92, 88.16, 92.65
    ), `2008` = c(NA, NA, 18.85, 71.21, 92.42, 77.59), `2007` = c(NA, 
    NA, 50.93, 79.69, 80.36, 78), delisted = c("NO", "NO", "NO", 
    "NO", "NO", "NO"), rank20 = c(535, 535, 646, 647, 648, 649
    ), rank19 = c(535, 535, 646, 647, 648, 649)), row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame"))

So essentially, I want to rank, take the top 30 companies, order them alphabetically to create a new dataframe with the names (column named "Name") of the companies per each year from 2007 to 2019. The end goal is to obtain the list per each year that displays the names of the companies ranked and filtered as above, in alphabetic order.

Answer 1

As @Parfait mentioned it becomes very easier to do data manipulation if you keep data in long format, you could do something like this:

library(dplyr)

result <- emiscore %>%
            tidyr::pivot_longer(cols = `2019`:`2007`, names_to = 'year') %>%
            group_by(year) %>%
            top_n(30, value)

This selects top 30 values for each year.