# Compound dataframe filtering with negated boolean expression
**URL:** https://discourse.julialang.org/t/compound-dataframe-filtering-with-negated-boolean-expression/24877
**Category:** General Usage
**Tags:** question, dataframes
**Created:** [June 3, 2019, 8:33am UTC](https://discourse.julialang.org/t/compound-dataframe-filtering-with-negated-boolean-expression/24877 "2019-06-03T08:33:07Z")
**Posts on this page:** 6
**Page:** 1
### Author:  [@notpeerreviewed](https://discourse.julialang.org/u/notpeerreviewed)
#### Post date: [June 3, 2019, 8:33am UTC](https://discourse.julialang.org/t/compound-dataframe-filtering-with-negated-boolean-expression/24877/1 "2019-06-03T08:33:07Z")
I’m trying to reproduce a compound filter but it involves negation of a two part specification.
I’d normally do something like this in R, using dplyr
`filter(!(Fuel == "Gas" & Field == "Maari"))`
In simple terms, I want to be able to filter my dataframe to exclude all instances where the combination of Fuel == Gas, and Field == Maari. I have been able to get a single negated boolean to work, but I can’t seem to get the negated compound expression to evaluate correctly.
I’ve tried this expression, and a few variations, but I keep getting an error saying there is no method matching !(::BitArray{1}). If I remove the ! then the expression returns the expected filtered result.
```julia
test[!((test.Fuel .== "Gas") .& (test.Field .== Symbol("Maari"))), :]
```
Can someone suggest how I might correctly negate the expression above?
Cheers
Jeff
---
### Author:  [@nilshg](https://discourse.julialang.org/u/nilshg)
#### Post date: [June 3, 2019, 8:37am UTC](https://discourse.julialang.org/t/compound-dataframe-filtering-with-negated-boolean-expression/24877/2 "2019-06-03T08:37:32Z")
I think you’re looking for `.! ` (I asked a similar question a while ago, search for negating boolean array)
---
### Author:  [@notpeerreviewed](https://discourse.julialang.org/u/notpeerreviewed)
#### Post date: [June 3, 2019, 8:40am UTC](https://discourse.julialang.org/t/compound-dataframe-filtering-with-negated-boolean-expression/24877/3 "2019-06-03T08:40:37Z")
You’re a champion! I can’t believe I didn’t think to try that.
It is quite a change moving from R to Julia.
Thanks so much.
Jeff
---
### Author:  [@nilshg](https://discourse.julialang.org/u/nilshg)
#### Post date: [June 3, 2019, 8:59am UTC](https://discourse.julialang.org/t/compound-dataframe-filtering-with-negated-boolean-expression/24877/4 "2019-06-03T08:59:37Z")
Glad that worked! It is and it isn’t to me - I’d say coming from Python is easier, but most often the issues you’ll experience are method errors down to using an inappropriate type. For me in R I seem to suffer from the same problems, albeit they are harder to diagnose as the type information is not great, different things have different indexing patterns eg.
I guess coming to a new language error messages are always cryptic to some extent, but having used both R and Julia in parallel over the past few months to me there’s no comparison in what’s easier to debug
---
### Author:  [@Iulian.Cioarca](https://discourse.julialang.org/u/Iulian.Cioarca)
#### Post date: [July 29, 2022, 12:25pm UTC](https://discourse.julialang.org/t/compound-dataframe-filtering-with-negated-boolean-expression/24877/5 "2022-07-29T12:25:29Z")
I have a similar usecase. Let’s take:
```julia
df=DataFrame(x=[1,2,3])
```
I tried to use: `subset(df, "x"=>x -> x.==1)` in order to make a simple filter.
How can I modify the logic to get as subset a dataframe containing the values 1 and 2?
I tried `subset(df, "x"=>x -> x.==[1,2])` but it gives an error. I also have to mention that the number of values I want to filter to is arbitrary, not necessarily 2 as in this example.
---
### Author:  [@rocco\_sprmnt21](https://discourse.julialang.org/u/rocco_sprmnt21)
#### Post date: [July 29, 2022, 12:50pm UTC](https://discourse.julialang.org/t/compound-dataframe-filtering-with-negated-boolean-expression/24877/6 "2022-07-29T12:50:47Z")
try with this
```julia
df=DataFrame(x=rand(1:10,10))
this=Set([1,2,3])
subset(df, :x=>ByRow(x->x∈this))
subset(df, :x=>x->x .∈ Ref(this))
```