R: drop columns from tibbles inside a function - r

Question

This is a followthrough of this topic. Here are my 3 tibbles:
dftest_tw <- structure(list(text = c("RT #BitMEXdotcom: A new high: US$500M turnover in the last 24 hours, over 80% of it on $XBTUSD. Congrats to the team and thank you to our u…",
"RT #Crowd_indicator: Thank you for this nice video, #Nicholas_Merten",
"RT #Crowd_indicator: Review of #Cindicator by DataDash: t.co/D0da3u5y3V"
), Tweet.id = c("896858423521837057", "896858275689398272", "896858135314538497"
), created.date = structure(c(17391, 17391, 17391), class = "Date"),
created.week = c(33, 33, 33)), .Names = c("text", "Tweet.id",
"created.date", "created.week"), row.names = c(NA, -3L), class = c("tbl_df",
"tbl", "data.frame"))
dftest1_tw <- dftest_tw
dftest2_tw <- dftest_tw
myUserList <- ls(,pattern = "_tw")
Following yesterday topic, I have the wanted result when running this:
library(tidyverse)
lst <- mget(myUserList) %>%
map2(myUserList, ~mutate(.data = .x, Twitter.name = .y)) %>%
list2env(lst, envir = .GlobalEnv)
I need to drop a few columns for each df. This do the job when running on one df:
select_(dftest_tw, quote(-text), quote(-Tweet.id), quote(-created.date))
It seems like I have a serious probelm when it comes to apply code to each member of a list. I can't find a way to apply it to all df when using lapply, or writing a function:
MySelect <- function(x){
select_(x, quote(-text), quote(-Tweet.id), quote(-created.date))
x
}
for(var in myUserList){MySelect(get(var))}
Thank you for your help.