I am trying to find the unique values of column
countmap(df[:column_name])
but getting error
ArgumentError: syntax df[column] is not supported use df[!, column] instead
DataFrame is a two-dimensional object so you cannot index into it using one index (what you attempt to with df[:column_name]). You either should get a property like:
countmap(df.column_name)
or if you prefer to use strings
countmap(df."column_name")
or if you want to use indexing then you have to use two indices: row selector and column selector so this:
countmap(df[:, :column_name])
or
countmap(df[!, :column_name])
works. The difference is that df[!, :column_name] accesses the column without copying it while df[:, :column_name] makes a copy. Also note that you can also use strings instead of Symbols here like df[!, "column_name"] or df[!, "column_name"].
All the rules of indexing into a data frame are described here.
Related
I got this code from elsewhere and I wondering if someone can explain what the square brackets are doing.
matrix1[i,] <- df[[1]][]
I am using this to assign values to a matrix and it works but I am not sure what exactly it's doing. What does the initial set of [[]] mean followed by another []?
This might help you understand a bit. You can copy and paste this code and see the differences between different ways of indexing using [] and $. The only thing I can't answer for you is the second empty set of square brackets, from my understanding that does nothing, unless a value is within those brackets.
#Retreives the first column as a data frame
mtcars[1]
#Retrieves the first column values only (three different methods of doing the same thing)
mtcars[,1]
mtcars[[1]]
mtcars$mpg
#Retrieves the first row as a data frame
mtcars[1,]
#I can use a second set of brackets to get the 4th value within the first column
mtcars[[1]][4]
mtcars$mpg[4]
The general function of [ is that of subsetting, which is well documented both in help (as suggested in comments), and in this piece. The rest of of my answer is heavily based on that source.
In fact, there are operators for subsetting in R; [[,[, and $.
The [ and $ are useful for returning the index and named position, respectfully, for example the first three elements of vector a = 1:10 may be subsetted with a[c(1,2,3)]. You can also negatively subset to remove elements, as a[-1] will remove the first index.
The $ operator is different in that it only takes element names as input, e.g. if your df was a dataframe with a column values, df$values would subset that column. You can achieve the same [, but only with a quoted name such as df["values"].
To answer more specifically, what does df[[1]][] do?
First, the [[-operator will return the 1st element from df, and the following empty [-operator will pull everything from that output.
I have the following code
#abnormal return
exp.ret <- lm((RET-rf)~mkt.rf+smb+hml, data=tesla[tesla$period=="estimation.period",])
tesla$abn.ret <- (tesla$RET-tesla$rf)-predict(exp.ret,tesla)
#CAR during event window
CAR <- sum(tesla$abn.ret[tesla$period=="event.period",])
First section runs fine, but second gets this error:
"Error in tesla$abn.ret[tesla$period == "event.period", ] :
incorrect number of dimensions
I know that the solution is to remove the last comma:
#CAR during event window
CAR <- sum(tesla$abn.ret[tesla$period=="event.period"])
Just wondering what is the right pedagogical way of understanding it, why do I need a comma in the end in some cases, but some not, when I'm filtering for only parts of the data frame.
$ sign, [[]] and [] have different meanings.
In short:
$ sign and [[]] subsets one column of a dataframe or one item of a list.
The output of a subsetted dataframe will be a vector, while the output of a subsetted list will be a variable the same class as the original item, which can be a dataframe, another list, etc...
It's important to note that $ doesn't accept a column index (only a column name) and that you cannot insert two column names/index after $ or inside [[]].
[] slices a dataframe or a list sorting out one or more elements.
the class of the output variable will be the same as the original variable.
if you slice a dataframe using [], the output will be a dataframe, the same applies for lists, etc...
In your specific case, you used $ sign to subset your variable. Then, you tried to slice this output from the subset action using [ , ], but it turned out that the output is a vector, and a vector has always only one dimension and an error was fired. You should slice your vector using [] (the output will be a vector) or [[]] (the output will be a vector with length = 1).
Possible ways to subset tesla as you wish:
tesla$abn.ret[tesla$period == "event.period"]
tesla[["abn.ret"]][tesla$period == "event.period"]
tesla[tesla$period == "event.period", "abn.ret"]
You would achieve the same result using tesla[["period"]] instead of tesla$period.
For some extra details/examples, refer to An introduction to R, published by CRAN.
I hope it helped you somehow..!
tesla$abn.ret is one-dimensional. Each comma separates a dimension, so yours implies 2 dimensions.
Alternatively you could run
tesla[tesla$period=="event.period", "abn.ret"]
And get the same results, since tesla is 2-d.
If you look at the documentation with command ?'[', you find that the default behaviour of syntax x[i] is to drop one dimension away.
If you want to disable the dropping of the dimension, you have explicitly to write x[i,drop=False].
I am trying to compare two lists of DateTime values in a Google Sheets. I want to highlight all DateTimes in the first list which are also present in the second list.
I already tried to use MATCH(), COUNTIF(), FILTER() together with COUNTA() or other approaches. However, although the values in both lists are basically copies of each other (just with some missing values in the second list), no matches will be returned. All are "true" DateTime values which can be formatted by using any of the date and time formatting options.
The MATCH example:
=ISNUMBER(MATCH(A2,INDIRECT("OTHERSHEET!$A$2:$A"),0))
I assume that there might be differences in the DateTime interpretation in the comparison why I also tried to use N() without success. I will always get an error like Did not find value '43297.75867' in MATCH evaluation. respectively FALSE after the ISNUMBER().
If I just do something like =N(A1)=N(OTHERSHEET!A1) with matching DateTimes, I get TRUE.
Same principle, but shorter:
=MATCH(A1,INDIRECT("OTHERSHEET!A:A"),0)
custom formula: =ISNUMBER(MATCH(A1,INDIRECT("OTHERSHEET!$A$1:$A"),0))
I have combined different matrix with same row size using cbind(). I observe that the column name of resulting matrix is not accessible using $.
Is there any specific reason for this? Thanks.
The $ operator does not work for matrices. Instead use matrix_name[,'column_name'] or matrix_name[,i], being i the column index.
I am using the tabular() function to produce tables in r (tables library).
I want to compute CI's from the data in the output (let mytable be the output from tabular()). Simple enough I thought, except when I go to call a value from the matrix, I get the error Error in mytable[1, i] - 1 : non-numeric argument to binary operator. I thought this was odd, as when I call up a particular cell of the matrix (where as.matrix returned true for mytable), for example mytable[1, i] for some i, I get an interger. I then do the as.list for mytable and get true also, so I am not sure what this means. I guess the tabular() function stores the results as a special kind of matrix.
I am only trying to pull out the mean,sdev, and n, which I am able to just by typing the cell location, for example mytable[1, i] would return an 86. However, when I try to call up the value in qt(.975,df=(mytable[1,i]-1)) for example, I get the error above. Not sure really how to approach this except to manually enter the values into another matrix (which I would like to avoid). Or, if I can compute CI's directly in the tabular() function that would work also. Cheers.
I shall quote for you the Value section of the documentation on the function ?tabular:
An object of S3 class "tabular". This is a matrix of mode list, whose
entries are computed summary values, with the following attributes:
rowLabels - A matrix of labels for the rows. This will have the same
number of rows as the main matrix, but may have multiple columns for
different nested levels of labels. If a label covers multiple rows, it
is entered in the first row, and NA is used to fill following rows.
colLabels - Like rowLabels, but labelling the columns.
table - The original table expression being displayed. A list of the
original format specifications are attached as a "fmtlist" attribute.
formats - A matrix of the same shape as the main result, containing NA
for default formatting, or an index into the format list.
As the documentation says, each element of the matrix is a list. If your tabular object is called tab type tab[1,1] and you should see a list containing one of your table values. If I wanted to modify that value, I would probably do something like:
tab[1,1]$term <- value
just like you would modify values in any other list.
Type attributes(tab) and you'll see the items listed above, containing a lot of the formatting information and row/col headers.