filtering data frame based on NA on multiple columns
filter rows based on two columns in r
r filter in list
dplyr filter not in
dplyr remove rows
filter na in r
I have the following data frame lets call it
df, with the following observations:
id type company 1 NA NA 2 NA ADM 3 North Alex 4 South NA NA North BDA 6 NA CA
I want to retain only the records which do not have NA in column "type" and "company".
id type company 3 North Alex NA North BDA
df_non_na <- df[!is.na(df$company) || !is.na(df$type), ]
But this did not work.
Thanks in advance
We can get the logical index for both columns, use
& and subset the rows.
df1[!is.na(df1$type) & !is.na(df1$company),] # id type company #3 3 North Alex #5 NA North BDA
rowSums on the logical matrix (
is.na(df1[-1])) to subset.
Data Wrangling Part 3: Basic and more advanced ways to filter rows, Now, i would want to filter this data-frame such that i only get values more How to order data frame rows according to vector with specific order using R? fruit_Frame<-data.frame(Fruit=NA, Cost=NA, Quantity=NA)[-1,]; . Often you may want to filter a Pandas dataframe such that you would like to keep the rows if values of certain column is NOT NA/NAN. We can use Pandas notnull() method to filter based on NA/NAN values of a column. # filter out rows ina . dataframe with column year values NA/NAN >gapminder_no_NA = gapminder[gapminder.year.notnull()] 4.
Using dplyr, you can also use the
library(dplyr) df_non_na <- df %>% filter_at(vars(type,company),all_vars(!is.na(.)))
Filtering R data-frame with multiple conditions, library(tidyverse) df <- tibble( ~col1, ~col2, ~col3, 1, 2, 3, 1, NA, 3, NA, 2, 3 ) How to replace NA values in a dataframe with Zero's ? It is simple R: Filter a data frame based on values in two columns. In the most recent assignment of the Computing for Data Analysis course we had to filter a data frame which contained N/A values in two columns to only return rows which had no N/A's. I started with a data frame that looked like this:
You need AND operator (&), not OR (|) I also strongly suggest the tidyverse approach by using the dplyr function filter() and the pipe operator %>%, from dplyr as well:
library(dplyr) df_not_na <- df %>% filter(!is.na(company) & !is.na(type))
How to remove NA values with dplyr::filter(), dplyr filter na filter_if dplyr dplyr conditional filter r select rows containing string filter data based on two columns in r. I have the following data frame lets call it df We use auto filters to filter subsets of data in Excel, but this filtering is limited to actual data. We can use Advanced Filter feature to filter multiple columns with different criteria. Understanding with an example will be easy. Open a workbook in Excel and copy below data into the sheet. You can also use your own dataset instead of this data.
you can use
filtering data frame based on NA on multiple columns, Filtering data is one of the very basic operation when you work with data. If you want to know more about 'how to select columns' please check this post I have You can get rid of them easily with 'is.na()' function, which would return TRUE if the dplyr - Plyr specialised for data frames: faster & with remote datastores One easy way to achieve this is through merging. If you have all the conditions in df_filter then you can do this: df_results = df_filter %>% left_join(df_all) improve this answer. edited May 5 '18 at 1:28. answered Mar 21 '18 at 17:59. 10 silver badges. 21 bronze badges.
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df [ complete.cases(df), ]?
- Or the previous with a single
df[!is.na(df$company) | !is.na(df$type), ]
- I think this will remove the case where "id" is NA
- Could also try
library(data.table) ; na.omit(setDT(df), cols = c("type", "company"))
- @ David, thanks for this
- Dena answered this, all I had to do was use "|" instead of "||",
- But 1
|doesn't give you your desired output.
- @user3875610 I get 6 rows from
df1[!is.na(df1$company) | !is.na(df1$type), ]
- @akrun Why can't we use
|here? and Why
&? I thought
&specifies only if both columns have
- @MAPK It iss a bit of reverse logic. I guess it is the same logic from deMorgans law
- That will eliminate rows that have any
NAvalues -- accepted answer already does the job, question has been resolved.