I want to create a column that codes for whether patients have had a comorbid diagnosis of depression or not. Problem is, the diagnosis can be recorded in one of 4 columns:
ComorbidDiagnosis;
OtherDiagnosis;
DischargeDiagnosis;
OtherDischargeDiagnosis.
I've been using
levels(dataframe$ynDepression)[levels(dataframe$ComorbidDiagnosis)=="Depression"]<-"Yes"
for all 4 columns but I don't know how to code those who don't have a diagnosis in any of the columns. I tried:
levels(dataframe$ynDepression)[levels(dataframe$DischOtherDiagnosis &
dataframe$OtherDiagnosis &
dataframe$ComorbidDiagnosis &
dataframe$DischComorbidDiagnosis)==""]<-"No"
I also tried using && instead but it didn't work. Am I missing something?
Thanks in advance!
Edit: I tried uploading an image of some example data but I don't have enough reputations to upload images yet. I'll try to put an example here but might not work:
Patient ID PrimaryDiagnosis OtherDiagnosis ComorbidDiagnosis
_________AN__________Depression
_________AN
_________AN__________Depression______PTSD
_________AN_________________________Depression
What's inside the [] must be (transformable to) a boolean for the subset to work. For example:
x<-1:5
x[x>3]
#4 5
x>3
# F F F T T
works because the condition is a boolean vector. Sometimes, the booleanship can be implicite, like in dataframe[,"var"] which means dataframe[,colnames(dataframe)=="var"] but R must be able to make it a boolean somehow.
EDIT : As pointed out by beginneR, you can also subset with something like df[,c(1,3)], which is numeric but works the same way as df[,"var"]. I like to see that kind of subset as implicit booleans as it enables a yes/no choice but you may very well not agree and only consider that they enable R to select columns and rows.
In your case, the conditions you use are invalid (dataframe$OtherDiagnosisfor example).
You would need something like rowSums(df[,c("var1","var2","var3")]=="")==3, which is a valid condition.
Related
So I kind of already know the possible solution but I don't know how to exactly go about it so please give me a bit of grace here.
I have a dataset for youtube trends that I want to read the values from two columns (likes and dislikes) and based off their contents I want an entry to be made in the new column. If the likes are higher than the dislikes I want it to be said as a 'positive' video and if it has more dislikes it should be 'negative'.
I'm primarily not sure how to go about this since most of the previous asks are based off of one column rather than two. I know some mentioned using cut, but would it still work the same?
all help is appreciated, thanks.
You can use a simple ifelse :
df$new_col <- ifelse(df$likes > df$dislikes, 'positive', 'negative')
This can also be written without ifelse as :
df$new_col <- c('negative', 'positive')[as.integer(df$likes > df$dislikes) + 1]
You can use Vectorize to create a vectorized version of a function. vfunc <- Vectorize(func) will allow you to call df$newcol <- vfunc(df$likes, df$dislikes) if your function takes two arguments and then return the result for each row in a vector that's assigned to a new column.
My first question here and I am not very experienced, however I hope this question is easy enough to answer since I only want to know if what I describe in the title is possible.
I have multiple dataframes taken from online capacity tests participants did.
For all Items I have response, score, and durationvariables among others.
Now I want to delete rows where all responsevariables are NA. So I can't just use a command to delete rows with where all is NA but there are also to many columns to do it by hand. And I also want to keep the dataframe together while doing it in order to really drop the complete rows, so just extracting all responsevariables doesn't sound like a good option.
However, besides a 3digit number based on the specific items the responsevariablenames are basically the same.
So instead of writing a very long impractical line mentioning all responsevariables and to drop the row if they all contain NA is there a way to not use the full anme of a variable but only use the end of the name for example so R checks the condition for all variables ending that way?
simplified e.g: instead of
newdf <- olddf[!(olddf$item123response != NA & olddf$item131response != NA & etc),]
Can I just do something like newdf <- olddf[!(olddf$xxxresponse != NA),] ?
I tried to google an answer but I didn't know how to frame my question effectively.
Thanks in advance!
Try This
newdf <- olddf[complete.cases(olddf[, grep('response', names(olddf))]), ]
I’m looking for a simple expression that puts a ‘1’ in column E if ‘SomeContent’ is contained in column D. I’m doing this in Azure ML Workbench through their Add Column (script) function. Here’s some examples they give.
row.ColumnA + row.ColumnB is the same as row["ColumnA"] + row["ColumnB"]
1 if row.ColumnA < 4 else 2
datetime.datetime.now()
float(row.ColumnA) / float(row.ColumnB - 1)
'Bad' if pd.isnull(row.ColumnA) else 'Good'
Any ideas on a 1 line script I could use for this? Thanks
Without really knowing what you want to look for in column 'D', I still think you can find all the information you need in the examples they give.
The script is being wrapped by a function that collects the value you calculate/provide and puts it in the new column. This assignment happens for each row individually. The value could be a static value, an arbitrary calculation, or it could be dependent on the values in the other columns for the specific row.
In the "Hint" section, you can see two different ways of obtaining the values from the other rows:
The current row is referenced using 'row' and then a column qualifier, for example row.colname or row['colname'].
In your case, you obtain the value for column 'D' either by row.D or row['D']
After that, all you need to do is come up with the specific logic for ensuring if 'SomeContent' is contained in column 'D' for that specific row. In your case, the '1 line script' would look something like this:
1 if [logic ensuring 'SomeContent' is contained in row.D] else 0
If you need help with the logic, you need to provide more specific examples.
You can read more in the Azure Machine Learning Documentation:
Sample of custom column transforms (Python)
Data Preparations Python extensions
Hope this helps
I did not find an answer to this - in my opinion quite basic - question. So in case I missed out on an already existing solution, I am sorry for that and would appreciate a link to the thread.
I am facing the following problem:
I want to create an if-condition whether or not an observations fulfills certain criteria. However, the set of variables i want to test is unknown, as they are created in the process and might change, depending on the data fed into the model.
I now have hard-coded the variable names, like below:
data$selectvar <- ifelse(data$crit1 == 1 | data$crit2 == 1 | data$crit3 == 1, 1, 0)
In above example, there could be cases where e.g., I only have crit1 and crit3 in the data set data, but not crit2. So the if condition would throw an error in these cases.
The way I have named the variables is that they all have the same prefix, so maybe there is a way to work with grepl or similar, but I don't know how.
There are many ways to do it, but if you want to use ifelse then you can try this..
ifelse(data[,grep("crit",colnames(data))]==1,1,0)
I am attempting to make elements in the first column of my df null (no entry at all) if it is equal to the element in the same row in the second column. This is a very simple thing, but I haven't been able to find the answer in the message boards.
Below are two of my attempts:
ifelse(y2014[y2014[,1]==y2014[,2]],y2014[,1]=="",y2014[,1]==y2014[,1])
y2014$new=ifelse(y2014[,1]==y2014[,2],0,y2014[,1])
Both give the following error: "level sets of factors are different" I checked the number of levels in each and they're equal, though several cells are blank in column 2. Would an apply function work better for what I'm trying to accomplish?
Really appreciate your help for a newbie.
Two things, factors generally need to be converted to character prior to comparing, and you want to assign NA rather than 0 to the value.
Something like this might be better:
y2014$new <- y2014[,1]
y2014$new[as.character(y2014$new) == as.character(y2014[,2])] <- NA