library(dplyr)library(zoo)# for the na.locf functiondata%>%group_by(ts_id)%>%# group by idmutate(value=na.locf(value,na.rm=F))# na.locf fills with the last non-empty value#head()# # A tibble: 6 x 3# # Groups: ts_id [2]# ts_id date value# <dbl> <date> <dbl>#...
1 r: replace_na() not replacing NAs 0 Replace NA values in a R data frame 0 How to use 'sparklyr::replace.na()' for replacing NaN on one column? 6 Why doesn't tidyr:fill replace my NA values 1 Replacing non-zero values while leaving na's as is 1 pro...
plyr::mapvalues可以像这样使用: mapvalues(mtcars$cyl, c(4, 6, 8), c("a", "b", "c")) 但这不起作用: mtcars %>% dplyr::select(cyl) %>% mapvalues(c(4, 6, 8), c("a", "b", "c")) %>% as.data.frame() 如何在dplyr中使用plyr::mapvalues?或者更好的是,dplyr的等价物是...
在第一次调用replace_na(df$n,0)时，第一个参数是一个向量（df$n），replace必须是一个值（标量）。然而，在第二个表达式df%>%replace_na(n,0)中，第一个data参数是一个框架，所以replace必须是一个命名列表，其中的名称指示要查找和替换NA值的列。我们将使用list(n=0)，因为您希望在名为n的列中将NA值...
This casual note is to record how to use R to replace the NA with 0 or any string. Generally, NA can be generated for different reasons, like unclean data, data transformation, or missing values. Othe
Using Manually Predicted Values The process of constructing a similar plot inggplot()follows the same general procedure as that for a simple linear regression. First, make a data frame that has the observed variables used in the model and predicted values and confidence limits for each observation...
替换R列值 、、 我把调查数据输入到一个数据框架中,我有一列值为P、G、S和N的列,我需要用1代替P,用2代替G,用3代替S,用4代替N。df1$type <- replace(as.numeric(df1$type), df1$type == "P", "1") 我试过使用这个,但是这会影响到所有的值,而不仅仅是P。 浏览2提问于2015-03-13得票数 0 ...
I'm curious if I could replace NA values in my data frame with text from the column to the left (that does not have NA), with an additional "_unclassified" text on the end. Here is an example data frame: feature<-c("1","2","3","4","5")phylum<-c("Firmic...
You want to replace the values in a dataset column, but you're getting an error like this: invalid factor level, NA generated Try this instead: levels(dataframe$column)[levels(dataframe$column)=='old_value'] <- 'new_value' Share Improve this answer Follow answered Jan 29, 2019 at 23...
The.xabove represents the element being evaluated (using a purrr-style functioning). This results in # A tibble: 5 x 5x y z a b<chr><chr><chr><chr><chr>1Bad Bad Bad Bad Bad2Bad Bad3Excellent333Bad3Excellent Good4Good Bad3Good Excellent5Excellent3Bad Excellent Bad ...