tidyr spread does not work
#> 5 2009-01-05 -0.5532382 -1.5550062 5.588266035 that there are two types of missingness in the input: explicit missing
(And it returns a data frame, not a matrix.) I finally figured out why I got the message "All columns must be named." If the class of the value column was factor or date, note that will not be true of the new columns that are produced, which are coerced to character before type conversion.
#> 7 2009-01-07 -0.7954609 0.8268879 2.950517792 #> 3 Z 0.001920052 3.0205002 1.3696140 0.6738913 5.5882660 -2.7163817
Featured on Meta Thanks for sharing that knowledge. Therefore, we use tidyr gather() and separate() functions to quickly tidy our data and reshape2dcast() to aggregate them. If you have struggled to understand what exactly these two functions do, this tutorial is for you.
install.packages ("tidyr") library (tidyr) %>% Operator. #> $ row : num 1 51 Then I tried to use the spread command to get to the 2nd table, but I got the following error message:Following on the comment from ayk, I'm providing an example. #> time X Y Z But this same approach doesn't work for the spread function. This is specific to a data_frame (note the dplyr class with the underscore in the name), as this works in my example when you have values of NA in a data.frame. Private self-hosted questions and answers for your enterpriseProgramming and related technical career opportunitiesThis is such good information to know. for egWhich is solved using match function which gives the required column positionBut this same approach doesn't work for the spread functionDo gather and spread functions take different column specification.
This is because in my actual dataset, I have NA inside and when I do the spread, it does not have a name for the column. tidyr isn't built-in, but you can see reshape2::dcast for a more flexible version of spread. I'm not sure if this is the exact reason why rows must be uniquely identifiable, but it makes sense. By using our site, you acknowledge that you have read and understand our In particular, existing methods only work for data frames, and tidyr never aggregates.
Looks like the spread does not recognize the n … This method does not work. This method does not work. View source: R/spread.R. However, when the n column on the first table (i.e, df_1) is system generated using the tally function. #> $ Species : chr "setosa" "versicolor"
I am trying to use spread and gather using tidyr version 1.0.0.9000 and dplyr version 0.8.3.9000. This method does not work.
8-)This method works in the code shown above. Error: Invalid column specification Do gather and spread functions take different column specification. The Overflow Blog #> 1 2009-01-01 -2.0524039 -1.4041640 0.001920052 You could modify the data directly to change the NAs to a more descriptive value. #> 10 2009-01-10 1.0199337 2.4345434 5.802172853#> stock 2009-01-01 2009-01-02 2009-01-03 2009-01-04 2009-01-05 2009-01-06 Is there a way I can pass in a variable that contains the column name
It looks to me like when you have a data_frame with a column of either a factor or character class that has values of NA, this cannot be spread without either removing them or re-classifying the data. If you want to spread multiple columns, you should mutate(paste) them into a single one, and spread them afterwards. Now this piece of code works fine when I do it standalone script. #> $ Species : chr "setosa" "versicolor" It’s designed specifically for tidying data, not general reshaping. Stack Overflow works best with JavaScript enabled Recommend：r - Reshaping data - is this an operation for tidyr::spread olumn. Thanks again for your help with this. 8-) It still gives me the Error: All columns must be named.
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