filtering data frame based on NA on multiple columns

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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

I tried:

 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

Or use rowSums on the logical matrix (is.na(df1[-1])) to subset.

df1[!rowSums(is.na(df1[-1])),]

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 filter_at function

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))

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you can use

na.omit(data_frame_name)

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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Manipulating data tables with dplyr, Selecting columns and filtering rows; Pipes the dplyr package; Be able to select subsets of columns from a dataframe, and filter rows according to a condition(s) Packages in R are basically sets of additional functions that let you do more stuff. In this hindfoot_half column, there are no NA values and all values are < 30. I have a data frame and tried to select only the observations I'm interested in by this: data[data["Var1"]>10] Unfortunately, this command destroys the data.frame structure and returns a long vector. What I want to get is the data.frame shortened by the observations that don't match my criteria.

Comments
  • df [ complete.cases(df), ] ?
  • Or the previous with a single | . ie: 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 NA.
  • @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 NA values -- accepted answer already does the job, question has been resolved.