Develop Reference

Pandas Apply returns a Series instead of a dataframe

The goal of following code is to go through each row in df_label, extract app1 and app2 names, filter df_all using those two names, concatenate the result and return it as a dataframe. Here is the code:
def create_dataset(se):
# extracting the names of applications
app1 = se.app1
app2 = se.app2
# extracting each application from df_all
df1 = df_all[df_all.workload == app1]
df1.columns = df1.columns + '_0'
df2 = df_all[df_all.workload == app2]
df2.columns = df2.columns + '_1'
# combining workloads to create the pairs dataframe
df3 = pd.concat([df1, df2], axis=1)
display(df3)
return df3
df_pairs = pd.DataFrame()
df_label.apply(create_dataset, axis=1)
#df_pairs = df_pairs.append(df_label.apply(create_dataset, axis=1))
I would like to append all dataframes returned from apply. However, while display(df3) shows the correct dataframe, when returned from function, it's not a dataframe anymore and it's a series. A series with one element and that element seems to be the whole dataframe. Any ideas what I am doing wrong?

When you select a single column, you'll get a Series instead of a DataFrame so df1 and df2 will both be series.
However, concatenating them on axis=1 should produce a DataFrame (whereas combining them on axis=0 would produce a series). For example:
df = pd.DataFrame({'a':[1,2],'b':[3,4]})
df1 = df['a']
df2 = df['b']
>>> pd.concat([df1,df2],axis=1)
a b
0 1 3
1 2 4
>>> pd.concat([df1,df2],axis=0)
0 1
1 2
0 3
1 4
dtype: int64

Fill in empty header with previous column name - pandas

I have a dataframe where each second column name is skipped:
eg
Step_1.
The idea is to fill unnamed columns with previous name to get:
Step_2.
To sum up "in" and "out" in each class, to get final result like this
The intermediary Step_1 is important and cannot be skipped to get the final result.
I appreciate any help and apologize for not being clear enough when asking question at the first attempt.
Thank you

Idea is convert columns to Series, so possible replace missing values instead values starting by Unnamed with forward filling:
df.columns = df.columns.to_series().mask(lambda x: x.str.startswith('Unnamed')).ffill()
print (df)
Column_1 Column_1 Column_2 Column_2
0 a d f g
EDIT:
If missing values in index:
df.columns = df.columns.to_series().ffill()
MultiIndex solution is necessary, if second row is header too - first use header=[0,1] for MultiIndex:
import pandas as pd
temp=u"""Column_1;Unnamed_column;Column_2;Unnamed_column
a;d;f;g
1;5;5;6
7;8;9;4"""
#after testing replace 'pd.compat.StringIO(temp)' to 'filename.csv'
df = pd.read_csv(pd.compat.StringIO(temp), sep=";", header=[0,1])
print (df)
Column_1 Unnamed_column Column_2 Unnamed_column
a d f g
0 1 5 5 6
1 7 8 9 4
a = df.columns.get_level_values(0)
b = df.columns.get_level_values(1)
df.columns = [a.to_series().mask(lambda x: x.str.startswith('Unnamed')).ffill(), b]
print (df)
Column_1 Column_2
a d f g
0 1 5 5 6
1 7 8 9 4

I tried this,
t = pd.DataFrame(df.columns)
t.loc[t[0].str.startswith('Unnamed: '),0] = np.NaN
t[0].bfill(inplace=True)
df.columns = t[0].values
Create temp dataframe with column of original dataframe. apply ffill or bfill as per your wish. assign back the values again to original dataframe.

You can rewrite the df.index with a list comprehension.
from itertools import chain
df = pd.DataFrame(
{"Column_1": [1], "Unnamed_column1": [2], "Column_2": [3], "Unnamed_column2": [4]})
cols = [[c, c] for c in df.columns[::2]]
df.columns = [_ for _ in chain(*cols)]
Having said that it might be better to assign unique names to columns as they will be used keys/indices, i.e .
cols = [[c, c+"_new"] for c in df.columns[::2]]

How to select dataframe row using regular expression on the index column?

I'm a pandas newbie.
Here's the problem with an example
df = pd.DataFrame(data={'id':['john','joe','zack']})
I know that I can select rows where the "id" column contains "jo" like so
mask = df['id'].str.contains('jo')
df[mask]
But suppose that id column is indexed
df = df.set_index('id')
Now how do I select the rows where the index column contains "jo"?

You need to change id to index:
df = pd.DataFrame(data={'id':['john','joe','zack'],
'col':[1,2,3]})
df = df.set_index('id')
df1 = df[df.index.str.contains('jo')]
print (df1)
col
id
john 1
joe 2

Pandas: get string from specific column header

Context:
i am new in Pandas and I need a function that creates new columns based on existing columns. The new columns name will have the name from the original column plus new characters (example: create a "As NEW" column from "As" column).
Can i access the old column header string to make the name of the new column?
Problem:
i have df['columnA'] and need to get "columnA" string

If I understand you correctly, this may be what you're looking for.
You can use str.contains() for the columns, then use string formatting to create the new column name.
df = pd.DataFrame({'col1':['A', 'A', 'B','B'], 'As': ['B','B','C','C'], 'col2': ['C','C','A','A'], 'col3': [30,10,14,91]})
col = df.columns[df.columns.str.contains('As')]
df['%s New' % col[0]] = 'foo'
print (df)
As col1 col2 col3 As New
0 B A C 30 foo
1 B A C 10 foo
2 C B A 14 foo
3 C B A 91 foo

Assuming that you have an empty DataFrame df with columns, you could access the columns of df as a list with:
>>> df.columns
Index(['columnA', 'columnB'], dtype='object')
.columns will allow you to overwrite the columns of df, but you don't need to pass in another Index. You can pass it a regular list, like so:
>>> df.columns = ['columna', 'columnb']
>>> df
Empty DataFrame
Columns: [columna, columnb]
Index: []

This can be done through the columns attribute.
cols = df.columns
# Do whatever operation you want on the list of strings in cols
df.columns = cols

pandas convert grouped rows into columns

I have a dataframe such as:
label column1
a 1
a 2
b 6
b 4
I would like to make a dataframe with a new column, with the opposite value from column1 where the labels match. Such as:
label column1 column2
a 1 2
a 2 1
b 6 4
b 4 6
I know this is probably very simple to do with a groupby command but I've been searching and can't find anything.

The following uses groupby and apply and seems to work okay:
x = pd.DataFrame({ 'label': ['a','a','b','b'],
'column1': [1,2,6,4] })
y = x.groupby('label').apply(
lambda g: g.assign(column2 = np.asarray(g.column1[::-1])))
y = y.reset_index(drop=True) # optional: drop weird index
print(y)

you can try the code block below:
#create the Dataframe
df = pd.DataFrame({'label':['a','a','b','b'],
'column1':[1,2,6,4]})
#Group by label
a = df.groupby('label').first().reset_index()
b = df.groupby('label').last().reset_index()
#Concat those groups to create columns2
df2 = (pd.concat([b,a])
.sort_values(by='label')
.rename(columns={'column1':'column2'})
.reset_index()
.drop('index',axis=1))
#Merge with the original Dataframe
df = df.merge(df2,left_index=True,right_index=True,on='label')[['label','column1','column2']]
Hope this helps

Assuming their are only pairs of labels, you could use the following as well:
# Create dataframe
df = pd.DataFrame(data = {'label' :['a', 'a', 'b', 'b'],
'column1' :[1,2, 6,4]})
# iterate over dataframe, identify matching label and opposite value
for index, row in df.iterrows():
newvalue = int(df[(df.label == row.label) & (df.column1 != row.column1)].column1.values[0])
# set value to new column
df.set_value(index, 'column2', newvalue)
df.head()

You can use groupby with apply where create new Series with back order:
df['column2'] = df.groupby('label')["column1"] \
.apply(lambda x: pd.Series(x[::-1].values)).reset_index(drop=True)
print (df)
column1 label column2
0 1 a 2
1 2 a 1
2 6 b 4
3 4 b 6