Imagine I have the following DataFrame :
10 rows x 26 columns named A to Z
What I would like to do is to make a multiple subset of the columns by their name (not the index). For instance, assume that I want columns A to D and P to Z in a new DataFrame named df2.
I tried something like this but it doesn't seem to work :
df2=df[:,[:A,:D ; :P,:Z]]
syntax: unexpected semicolon in array expression
top-level scope at Slicing.jl:1
Any idea of the way to do it ?
Thanks for any help
df2 = select(df, Between(:A,:D), Between(:P,:Z))
or
df2 = df[:, All(Between(:A,:D), Between(:P,:Z))]
if you are sure your columns are only from :A to :Z you can also write:
df2 = select(df, Not(Between(:E, :O)))
or
df2 = df[:, Not(Between(:E, :O))]
Finally, you can easily find an index of the column using columnindex function, e.g.:
columnindex(df, :A)
and later use column numbers - if this is something what you would prefer.
In Julia you can also build Ranges with Chars and hence when your columns are named just by single letters yet another option is:
df[:, Symbol.(vcat('A':'D', 'P':'Z'))]
Related
I have two dataframes.
df1 has an index list made of strings like (row1,row2,..,rown) and a column list made of strings like (col1,col2,..,colm) while df2 has k rows and 3 columns (char_1,char_2,value). char_1 contains strings like df1 indexes while char_2 contains strings like df1 columns. I only want to assign the df2 value to df1 in the right position. For example if the first row of df2 reads ['row3','col1','value2'] I want to assign value2 to df1 in the position ([2,0]) (third row and first column).
I tried to use two functions to slide rows and columns of df1:
def func1(val):
# first I convert the series to dataframe
val=val.to_frame()
val=val.reset_index()
val=val.set_index('index') # I set the index so that it's the right column
def func2(val2):
try: # maybe the combination doesn't exist
idx1=list(cou.index[df2[char_2]==(val2.name)]) #val2.name reads col name of df1
idx2=list(cou.index[df2[char_1]==val2.index.values[0]]) #val2.index.values[0] reads index name of df1
idx= list(reduce(set.intersection, map(set, [idx1,idx2])))
idx=int(idx[0]) # final index of df2 where I need to take value to assign to df1
check=1
except:
check=0
if check==1: # if index exists
val2[0]=df2['value'][idx] # assign value to df1
return val2
val=val.apply(func2,axis=1) #apply the function for columns
val=val.squeeze() #convert again to series
return val
df1=df1.apply(func1,axis=1) #apply the function for rows
I made the conversion inside func1 because without this step I wasn't able to work with series keeping index and column names so I wasn't able to find the index idx in func2.
Well the problem is that it takes forever. df1 size is (3'600 X 20'000) and df2 is ( 500 X 3 ) so it's not too much. I really don't understand the problem.. I run the code for the first row and column to check the result and it's fine and it takes 1 second, but now for the entire process I've been waiting for hours and it's still not finished.
Is there a way to optimize it? As I wrote in the title I only need to run a function that keeps column and index names and works sliding the entire dataframe. Thanks in advance!
I want to add a new column to a datframe "table" (name: conc) which uses the values in columns (plate, ab) to get the numeric value from the dataframe "concs"
Below is what I mean, with the dataframe "exp" used to show what I expect the data to look like
what is the proper way to do this. Is it using some multiple condition, or do I need to reshape the concs dataframe somehow?
Use DataFrame.melt with left join for new column concs, if no match is created NaNs:
exp = concs.melt('plate', var_name='ab', value_name='concs').merge(table,on=['plate', 'ab'], how='left')
Solution should be simplify - if same columns names 'plate', 'ab' in both DataFrames and need merge by both is possible omit on parameter:
exp = concs.melt('plate', var_name='ab', value_name='concs').merge(table, how='left')
First melt the concs dataframe and then merge with table:
out = concs.melt(id_vars=['plate'],
value_vars=concs.columns.drop('plate').tolist(),
var_name='ab').merge(table, on=['plate', 'ab'
]).rename(columns={'value': 'concs'})
or just make good use of parameters of melt like in jezraels' answer:
out = concs.melt(id_vars=['plate'],
value_name='concs',
var_name='ab').merge(table, on=['plate', 'ab'])
I have 2 dataframes, the first one has 53 columns and the second one has 132 column.
I want to compare the 2 dataframes and remove all the columns that are not in common between the 2 dataframes and then display each dataframe containing only those columns that are common.
What I did so far is to get a list of all the column that dont't match, but I don't know how to drop them.
val diffColumns = df2.columns.toSet.diff(df1.columns.toSet).union(df1.columns.toSet.diff(df2.columns.toSet))
This is getting me a scala.collection.immutable.Set[String].
Now I'd like to use this to drop these columns from each dataframe. Something like that, but this is not working...
val newDF1 = df1.drop(diffColumns)
The .drop function accepts a list of columns, not the Set object, so you need to convert it to Seq and "expand it" using, the : _* syntax, like, this:
df.drop(diffColumns.columns.toSet.toSeq: _*)
Also, instead of generating diff, it could be just easier to do intersect to find common columns, and use .select on each dataframe to get the same columns:
val df = spark.range(10).withColumn("b", rand())
val df2 = spark.range(10).withColumn("c", rand())
val commonCols = df.columns.toSet.intersect(df2.columns.toSet).toSeq.map(col)
df.select(commonCols: _*)
df2.select(commonCols: _*)
I am extracting tables from pdf using Camelot. Two of the columns are getting merged together with a newline separator. Is there a way to separate them into two columns?
Suppose the column looks like this.
A\nB
1\n2
2\n3
3\n4
Desired output:
|A|B|
|-|-|
|1|2|
|2|3|
|3|4|
I have tried df['A\nB'].str.split('\n', 2, expand=True) and that splits it into two columns however I want the new column names to be A and B and not 0 and 1. Also I need to pass a generalized column label instead of actual column name since I need to implement this for several docs which may have different column names. I can determine such column name in my dataframe using
colNew = df.columns[df.columns.str.contains(pat = '\n')]
However when I pass colNew in split function, it throws an attribute error
df[colNew].str.split('\n', 2, expand=True)
AttributeError: DataFrame object has no attribute 'str'
You can take advantage of the Pandas split function.
import pandas as pd
# recreate your pandas series above.
df = pd.DataFrame({'A\nB':['1\n2','2\n3','3\n4']})
# first: Turn the col into str.
# second. split the col based on seperator \n
# third: make sure expand as True since you want the after split col become two new col
test = df['A\nB'].astype('str').str.split('\n',expand=True)
# some rename
test.columns = ['A','B']
I hope this is helpful.
I reproduced the error from my side... I guess the issue is that "df[colNew]" is still a dataframe as it contains the indexes.
But .str.split() only works on Series. So taking as example your code, I would convert the dataframe to series using iloc[:,0].
Then another line to split the column headers:
df2=df[colNew].iloc[:,0].str.split('\n', 2, expand=True)
df2.columns = 'A\nB'.split('\n')
I have a dataframe as
df = pd.DataFrame(np.random.randn(5,4),columns=list('ABCD'))
I can use the following to achieve the traditional calculation like mean(), sum()etc.
df.loc['calc'] = df[['A','D']].iloc[2:4].mean(axis=0)
Now I have two questions
How can I apply a formula (like exp(mean()) or 2.5*mean()/sqrt(max()) to column 'A' and 'D' for rows 2 to 4
How can I append row to the existing df where two values would be mean() of the A and D and two values would be of specific formula result of C and B.
Q1:
You can use .apply() and lambda functions.
df.iloc[2:4,[0,3]].apply(lambda x: np.exp(np.mean(x)))
df.iloc[2:4,[0,3]].apply(lambda x: 2.5*np.mean(x)/np.sqrt(max(x)))
Q2:
You can use dictionaries and combine them and add it as a row.
First one is mean, the second one is some custom function.
ad = dict(df[['A', 'D']].mean())
bc = dict(df[['B', 'C']].apply(lambda x: x.sum()*45))
Combine them:
ad.update(bc)
df = df.append(ad, ignore_index=True)