Is there a way to read in a selection of non-consecutive columns of Excel data using XLSX.gettable? I’ve read the documentation here XLSX.jl Tutorial, but it’s not clear whether it’s possible to do this. For example,
df = DataFrame(XLSX.gettable(sheet,"A:B")...)
selects the data in columns “A” and “B” of a worksheet called sheet. But what if I want columns A and C, for example? I tried
df = DataFrame(XLSX.gettable(sheet,["A","C"])...)
and similar variations of this, but it throws the following error: MethodError: no method matching gettable(::XLSX.Worksheet, ::Array{String,1}).
Is there a way to make this work with gettable, or is there a similar function which can accomplish this?
I don't think this is possible with the current version of XLSX.jl:
If you look at the definition of gettable here you'll see that it calls
eachtablerow(sheet, cols;...)
which is defined here as accepting Union{ColumnRange, AbstractString} as input for the cols argument. The cols argument itself is converted to a ColumnRange object in the eachtablerow function, which is defined here as:
struct ColumnRange
start::Int # column number
stop::Int # column number
function ColumnRange(a::Int, b::Int)
#assert a <= b "Invalid ColumnRange. Start column must be located before end column."
return new(a, b)
end
end
So it looks to me like only consecutive columns are working.
To get around this you should be able to just broadcast the gettable function over your column ranges and then concatenate the resulting DataFrames:
df = reduce(hcat, DataFrame.(XLSX.gettable.(sheet, ["A:B", "D:E"])))
I found that to get #Nils Gudat's answer to work you need to add the ... operator to give
reduce(hcat, [DataFrame(XLSX.gettable(sheet, x)...) for x in ["A:B", "D:E"]])
Related
I am having a problem with filtering a pandas dataframe. I am trying to filter a dataframe based on column values being equal to a specific list but I am getting a length error.
I tried every possible way of filtering a dataframe but got nowhere. Any help would be appreciated, thanks in advance.
Here is my code :
for ind in df_hourly.index:
timeslot = df_hourly['date_parsed'][ind][0:4] # List value to filter
filtered_df = df.loc[df['timeslot'] == timeslot]
Error : ValueError: ('Lengths must match to compare', (5696,), (4,))
Above Image : df , Below Image : df_hourly
In the above image, the dataframe I want to filter is shown. Specifically, I want to filter according to the "timeslot" column.
And the below image shows the the dataframe which includes the value I want to filter by. I specifically want to filter by "date_parsed" column. In the first line of my code, I iterate through every row in this dataframe and assign the first 4 elements of the list value in df_hourly["date_parsed"] to a variable and later in the code, I try to filter the above dataframe by that variable.
When comparing columns using ==, pandas try to compare value by value - aka does the first item equals to first item, second item to the second and so on. This is why you receive this error - pandas expects to have two columns of the same shape.
If you want to compare if value is inside a list, you can use the .isin (documentation):
df.loc[df['timeslot'].isin(timeslot)]
Depends on what timeslot is exactly, you might to take timeslot.values or something like that (hard to understand exactly without giving an example for your dataframe)
I am attempting to add a column to a dataframe, using a value from a specific column—-let’s assume it’s an id—-to look up its actual value from another df.
So I set up a lookup def
def lookup(id:String): String {
return lookupdf.select(“value”)
.where(s”id = ‘$id’”).as[String].first
}
The lookup def works if I test it on its own by passing an id string, it returns the corresponding value.
But I’m having a hard time finding a way to use it within the “withColumn” function.
dataDf
.withColumn(“lookupVal”, lit(lookup(col(“someId”))))
It properly complains that I’m passing in a column, instead of the expected string, the question is how do I give it the actual value from that column?
You cannot access another dataframe from withColumn . Think of withColumn can only access data at a single record level of the dataDf
Please use a join like
val resultDf = lookupDf.select(“value”,"id")
.join(dataDf, lookupDf("id") == dataDf("id"), "right")
Working with Julia 1.1:
The following minimal code works and does what I want:
function test()
df = DataFrame(NbAlternative = Int[], NbMonteCarlo = Int[], Similarity = Float64[])
append!(df.NbAlternative, ones(Int, 5))
df
end
Appending a vector to one column of df. Note: in my whole code, I add a more complicated Vector{Int} than ones' return.
However, #code_warntype test() does return:
%8 = invoke DataFrames.getindex(%7::DataFrame, :NbAlternative::Symbol)::AbstractArray{T,1} where T
Which means I suppose, thisn't efficient. I can't manage to get what this #code_warntype error means. More generally, how can I understand errors returned by #code_warntype and fix them, this is a recurrent unclear issue for me.
EDIT: #BogumiłKamiński's answer
Then how one would do the following code ?
for na in arr_nb_alternative
#show na
for mt in arr_nb_montecarlo
println("...$mt")
append!(df.NbAlternative, ones(Int, nb_simulations)*na)
append!(df.NbMonteCarlo, ones(Int, nb_simulations)*mt)
append!(df.Similarity, compare_smaa(na, nb_criteria, nb_simulations, mt))
end
end
compare_smaa returns a nb_simulations length vector.
You should never do such things as it will cause many functions from DataFrames.jl to stop working properly. Actually such code will soon throw an error, see https://github.com/JuliaData/DataFrames.jl/issues/1844 that is exactly trying to patch this hole in DataFrames.jl design.
What you should do is appending a data frame-like object to a DataFrame using append! function (this guarantees that the result has consistent column lengths) or using push! to add a single row to a DataFrame.
Now the reason you have type instability is that DataFrame can hold vector of any type (technically columns are held in a Vector{AbstractVector}) so it is not possible to determine in compile time what will be the type of vector under a given name.
EDIT
What you ask for is a typical scenario that DataFrames.jl supports well and I do it almost every day (as I do a lot of simulations). As I have indicated - you can use either push! or append!. Use push! to add a single run of a simulation (this is not your case, but I add it as it is also very common):
for na in arr_nb_alternative
#show na
for mt in arr_nb_montecarlo
println("...$mt")
for i in 1:nb_simulations
# here you have to make sure that compare_smaa returns a scalar
# if it is passed 1 in nb_simulations
push!(df, (na, mt, compare_smaa(na, nb_criteria, 1, mt)))
end
end
end
And this is how you can use append!:
for na in arr_nb_alternative
#show na
for mt in arr_nb_montecarlo
println("...$mt")
# here you have to make sure that compare_smaa returns a vector
append!(df, (NbAlternative=ones(Int, nb_simulations)*na,
NbMonteCarlo=ones(Int, nb_simulations)*mt,
Similarity=compare_smaa(na, nb_criteria, nb_simulations, mt)))
end
end
Note that I append here a NamedTuple. As I have written earlier you can append a DataFrame or any data frame-like object this way. What "data frame-like object" means is a broad class of things - in general anything that you can pass to DataFrame constructor (so e.g. it can also be a Vector of NamedTuples).
Note that append! adds columns to a DataFrame using name matching so column names must be consistent between the target and appended object.
This is different in push! which also allows to push a row that does not specify column names (in my example above I show that a Tuple can be pushed).
when print a pandas dataframe, how to print the first n row by default?
I find myself frequently doing df.head(10) to view the column names and first a couple of rows.
I prefer when I type 'df', it prints the first n row by default, instead of printing the whole df, which in this case I cannot see the column names.
If I understand you correctly, you may set
pd.options.display.max_rows = 10
and whenever you do just df in your notebook, only 10 rows would be displayed.
You can always set back to the default value doing
pd.reset_option('display.max_rows')
Check pd.describe_option('display') for more information
Curry DataFrame.head using functools.partial.
from functools import partial
head10 = partial(pd.DataFrame.head, n=10)
Now you can either call the function passing your DataFrame as an argument,
head10(df)
Or, pass the function to df.pipe (which internally passes df as an argument to your function),
df.pipe(head10)
To get the first 10 rows by default.
The other option is to create a new class that extends DataFrame and add your own function (e.g., headXX) which internally calls df.head(n=10) and returns the result.
See the subclassing DataFrame section in the docs.
I have a function that has multiple inputs, and would like to use SFrame.apply to create a new column. I can't find a way to pass two arguments into SFrame.apply.
Ideally, it would take the entry in the column as the first argument, and I would pass in a second argument. Intuitively something like...
def f(arg_1,arg_2):
return arg_1 + arg_2
sf['new_col'] = sf.apply(f,arg_2)
suppose the first argument of function f is one of the column.
Say argcolumn1 in sf, then
sf['new_col'] = sf['argcolumn1'].apply(lambda x:f(x,arg_2))
should work
Try this.
sf['new_col'] = sf.apply(lambda x : f(arg_1, arg_2))
The way i understand your question (and because none of the previous answers are marked as accepted), it seems to me that you are trying to apply a transformation using two different columns of a single SFrame, so:
As specified in the online documentation, the function you pass to the SFrame.apply method will be called for every row in the SFrame.
So you should rewrite your function to receive a single argument representing the current row, as follow:
def f(row):
return row['column_1'] + row['column_2']
sf['new_col'] = sf.apply(f)