tidyverse

问题 Some time ago everything was ok. But after recent updates (I cannot track down which ones), in RStudio the "Attaching packages" section of the tidyverse startup message has encoding issues while the "Conflicts" part is correct. In R Gui this issue is not present (and there are no different colors). How can I get a non-distorted tidyverse startup message in RStudio? (With v and not <U+2713> ). Is it a bug in some package or RStudio? Are there some settings incorrect on my side? Rstudio version

问题 I'm having trouble figuring out how to effective map across multiple parameters and variables within a tbl to generate new variables. In the "real" version, I basically have one mathematical function generating a central estimate, and I need to run a whole series of sensitivity tests varying different parameters. I'm trying to figure out how to do this within the tidyverse. It looks like map() and mutate() are the answers to this, but I'm having trouble. # building the practice dataset pracdf

问题 Case1: Input ST_DATE ND_DATE LO_NO ACTV_CODE ACTV_AMT AB_NO FEATURE_CODE L_NU 7/27/16 7/27/16 265 O 15 1 INTEREST 855 7/27/16 7/27/16 265 O 14 1 INTEREST 855 Expected output ST_DATE ND_DATE LO_NO ACTV_CODE ACTV_AMT AB_NO FEATURE_INTEREST L_NU 7/27/16 7/27/16 265 O 29 1 2 855 Case2: Input (my code is working for case2 but throwing error for case1) ST_DATE ND_DATE LO_NO ACTV_CODE ACTV_AMT AB_NO FEATURE_CODE L_NU 7/27/16 7/27/16 265 O 15 1 INTEREST 855 7/27/16 7/27/16 265 O 14 1 INSTALLMENT 855

问题 I've set up hosted RStudio on Ubunto and have loaded several packages already with no issues, including caret and lubridate. However, when I tried to install tidyverse I get: > install.packages("tidyverse", dependencies = T) Installing package into ‘/home/rstudio/R/x86_64-pc-linux-gnu-library/3.4’ (as ‘lib’ is unspecified) trying URL 'https://cloud.r-project.org/src/contrib/tidyverse_1.2.1.tar.gz' Content type 'application/x-gzip' length 61647 bytes (60 KB) ===================================

问题 This question already has an answer here : How to spread columns with duplicate identifiers? (1 answer) Closed 2 years ago . I have a problem with a dataframe that I need to reshape. I have this command: library(tidyverse) df1 = df1 %>% gather(Day, value, Day01:Day31) %>% spread(Station, value) And I get this error: Error: Duplicate identifiers for rows (130933, 131029), (389113, 389209), (647293, 647389), (905473, 905569), (1163653, 1163749), (1421833, 1421929), (1680013, 1680109), (1938193,

问题 I have the following process which uses group_split of dplyr: library(tidyverse) set.seed(1) iris %>% sample_n(size = 5) %>% group_by(Species) %>% group_split() The result is: [[1]] # A tibble: 2 x 5 Sepal.Length Sepal.Width Petal.Length Petal.Width Species <dbl> <dbl> <dbl> <dbl> <fct> 1 5 3.5 1.6 0.6 setosa 2 5.1 3.8 1.5 0.3 setosa [[2]] # A tibble: 2 x 5 Sepal.Length Sepal.Width Petal.Length Petal.Width Species <dbl> <dbl> <dbl> <dbl> <fct> 1 5.9 3 4.2 1.5 versicolor 2 6.2 2.2 4.5 1.5

问题 I have annual financial data for several stocks. I needed to blow it out to become monthly data and, thanks to an answer to this question I'd asked earlier, I have a solution which involves mutating the date column into lists of dates : library(tidyverse) library(lubridate) factors.subset.raw = structure(list( sec_id = c(1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1572L, 1676L, 1676L, 1676L,

问题 I am struggling with renumbering result from intersection/ group_indices in R for a few days. Sample data frame is shown below: t <- data.frame(mid=c(102,102,102,102,102,102,102,103,103,103,103,103,103,103), aid=c(10201,10202,10203,10204,10205,10206,10207, 10301,10302,10303,10304,10305,10306,10307), dummy=c(0,1,0,1,0,1,0,0,1,0,1,0,1,0), location=c(0,2,0,4,0,1,0,0,2,0,2,0,3,0) ) I need to update numbers stored in "location" fiels to sequential number by a group of "mid" without changing its

问题 My data looks like this: df <- tribble( ~A, ~B, 0.2, 0.1, 0.2, 0.3, 0.5, 0.1, 0.7, 0.9, 0.8, 0.9, 0.4, 0.2) How might I select the max/min value between A and B ? Desired Output: A B C 1 0.2 0.1 0.2 2 0.2 0.3 0.3 3 0.5 0.1 0.5 4 0.7 0.9 0.9 5 0.8 0.9 0.9 6 0.4 0.2 0.4 回答1: You could try pmax mutate(df, C=pmax(A,B)) # A B C #1 0.2 0.1 0.2 #2 0.2 0.3 0.3 #3 0.5 0.1 0.5 #4 0.7 0.9 0.9 #5 0.8 0.9 0.9 #6 0.4 0.2 0.4 max gets you the maximum single value of the two columns instead of the "row"

问题 I have a data frame. In my data set. a, a1 and a2 are the exact same variable. However when you have the same names in r it automatically adds a 1 at the end of the name. df = data.frame(a = rnorm(4), b = rnorm(4), c = rnorm(4), a1 = rnorm(4), b1 = rnorm(4), c1 = rnorm(4), a2 = rnorm(4), b2 = rnorm(4), c2 = rnorm(4), date = seq(as.Date("2019-05-05"),as.Date("2019-05-08"), 1)) print(df) a b c a1 b1 a2 b2 c2 date 1 -1.0938097 1.3948486 1.2805904 1.6187439 1.0200681 -1.4335761 -0.4583526 0