Rename columns of dataframe by days in R - r

Question

I need to rename a dataframe by days in analysis.
names(dados) <- c("name", "day_1","Freq_1","Percent_1","day_2","Freq_2","Percent_2",
"day_3","Freq_3","Percent_3","day_4","Freq_4","Percent_4",
"day_5","Freq_5","Percent_5","day_6","Freq_6","Percent_6",
"day_7","Freq_7","Percent_7","day_8","Freq_8","Percent_8",
"day_9","Freq_9","Percent_9")
I'm doing an analysis that the data I get is in a list of dataframes, where each dataframe represents a day of analysis. I combine the dataframes and I have the columns 'name' unique and 'day_X', 'Freq_X' and 'Percent_X' for each dataframe as a return.
As return I need the columns to have the following names:
"name", "day_1","Freq_1","Percent_1","day_2","Freq_2","Percent_2","day_3","Freq_3","Percent_3"
How do I go about analyzing 50 days?
reproducible example:
day1 <- data.frame(name = c("jose", "mary", "julia"), freq = c(1,5,3), percent = c(40,30,20))
day2 <- data.frame(name = c("abner", "jose", "mary"), freq = c(3,5,4), percent = c(20,30,20))
day3 <- data.frame(name = c("abner", "jose", "mike"), freq = c(6,2,3), percent = c(40,30,70))
day4 <- data.frame(name = c("andre", "joseph", "ana"), freq = c(1,5,8), percent = c(40,30,20))
day5 <- data.frame(name = c("abner", "poli", "joseph"), freq = c(4,3,3), percent = c(10,30,10))
dates <- list(day1,day2,day4,day5)
data <- Reduce(function(x, y) merge(x, y, by = "name", all = TRUE), dates)

Answer 1

Here's a way to get what you want using the tidyverse suite of packages. We start by putting the data in the "long" format - but add a column with the date:
long_form <- dates %>%
imap_dfr(function(x, y) dplyr::mutate(x, day_num = y))
Now, to get the wide format you are after, we need to reformat things a bit, as done in the following code. I'm not sure what is supposed to go in the day_# variables, as #useR mentioned in the comments, so it's missing. If you have a variable called day, the code should automatically do the right thing as written.
wide_form <- long_form %>%
gather(key, value, -name,-day_num) %>%
dplyr::mutate(
key = paste(key, day_num, sep = "_")
) %>%
select(-day_num) %>%
spread(key, value)

Answer 2

One can use dplyr::bind_rows to merge all data frames form the list to a data frame. Please provide name to list so that day1, day2 etc can set beforehand. Finally, gather and spread is used to transform the data.
names(dates) <- paste("day", seq_along(dates), sep = "")
library(tidyverse)
bind_rows(dates,.id = "Name") %>%
group_by(Name) %>%
mutate(rn = row_number()) %>%
ungroup() %>%
gather(Key, value, -Name,-rn) %>%
unite("Key", c("Key", "Name")) %>%
spread(Key, value) %>%
select(-rn)
Result:
# # A tibble: 3 x 12
# freq_day1 freq_day2 freq_day3 freq_day4 name_day1 name_day2 name_day3 name_day4 percent_day1 percent_day2 percent~ percent~
# * <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
# 1 1 3 1 4 jose abner andre abner 40 20 40 10
# 2 5 5 5 3 mary jose joseph poli 30 30 30 30
# 3 3 4 8 3 julia mary ana joseph 20 20 20 10
#
Data:
Data is slightly modified from OP. I have included stringsAsFactors = FALSE argument as part of data.frame to avoid a mutate_at call to convert factor to character.
day1 <- data.frame(name = c("jose", "mary", "julia"), freq = c(1,5,3), percent = c(40,30,20), stringsAsFactors = FALSE)
day2 <- data.frame(name = c("abner", "jose", "mary"), freq = c(3,5,4), percent = c(20,30,20), stringsAsFactors = FALSE)
day3 <- data.frame(name = c("abner", "jose", "mike"), freq = c(6,2,3), percent = c(40,30,70), stringsAsFactors = FALSE)
day4 <- data.frame(name = c("andre", "joseph", "ana"), freq = c(1,5,8), percent = c(40,30,20), stringsAsFactors = FALSE)
day5 <- data.frame(name = c("abner", "poli", "joseph"), freq = c(4,3,3), percent = c(10,30,10), stringsAsFactors = FALSE)
dates <- list(day1,day2,day4,day5)