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How to get the name of a column at a specific index and print it into another column using python and pandas

I'm a python newbie and need help with a specfic task. My main goal is to identify all indicies with their specific values and column-names which are greater than 0 within a row and to sum up these values below each other into another column within the same row.
Here is what I tried:
import pandas as pd
import numpy as np
table = {
'A':[0, 2, 0, 5],
'B' :[4, 1, 3, 0],
'C':[2, 9, 0, 6],
'D':[1, 0, 1, 6]
}
df = pd.DataFrame(table)
print(df)
# create a new column that sums up the row
df['summary'] = 'NoData'
# print the header
print(df.columns.values)
A B C D summary
0 0 4 2 1 NoData
1 2 1 9 0 NoData
2 0 3 0 1 NoData
3 5 0 6 6 NoData
# get length of rows and columns
row = len(df.index)
column = len(df.columns)
# If a value at a spefic index is greater
# than 0, take the column name and the value at that index and print it into the column
#'summary'. Also write all values greater than 0 within a row below each other
for i in range(row):
for j in range(column):
if df.iloc[i][j] > 0:
df.at[i,'summary'] = df.columns(df.iloc[i][j]) + '\n'
I hope it is a bit clear what I want to achieve. Here is a picture of how the result should look in the column 'summary'

You don't really need a for loop.
Starting with df:
A B C D
0 0 4 2 1
1 2 1 9 0
2 0 3 0 1
3 5 0 6 6
You can do:
# Define an helper function
def f(val, col_name):
# You can modify this function in order to customize the summary string
return "" if val == 0 else str(val) + col_name + "\n"
# Assign summary column
df["summary"] = df.apply(lambda x: x.apply(f, args=(x.name,))).sum(axis=1).str[:-1]
Output:
A B C D summary
0 0 4 2 1 4B\n2C\n1D
1 2 1 9 0 2A\n1B\n9C
2 0 3 0 1 3B\n1D
3 5 0 6 6 5A\n6C\n6D
It works for longer column names as well:
one two three four summary
0 0 4 2 1 4two\n2three\n1four
1 2 1 9 0 2one\n1two\n9three
2 0 3 0 1 3two\n1four
3 5 0 6 6 5one\n6three\n6four

Try this:
import pandas as pd
import numpy as np
table = {
'A':[0, 2, 0, 5],
'B' :[4, 1, 3, 0],
'C':[2, 9, 0, 6],
'D':[1, 0, 1, 6]
}
df = pd.DataFrame(table)
print(df)
print(f'\n\n-------------BREAK-----------\n\n')
def func(line):
templist = ''
list_col = line.index.values.tolist()
temp = line.values.tolist()
for x in range(0, len(temp)):
if (temp[x] <= 0):
pass
else:
if (x == 0 ):
templist = f"{temp[x]}{list_col[x]}"
else:
templist = f"{templist}\n{temp[x]}{list_col[x]}"
return templist
df['summary'] = df.apply(func, axis = 1)
print(df)
EXIT
A B C D
0 0 4 2 1
1 2 1 9 0
2 0 3 0 1
3 5 0 6 6
-------------BREAK-----------
A B C D summary
0 0 4 2 1 \n4B\n2C\n1D
1 2 1 9 0 2A\n1B\n9C
2 0 3 0 1 \n3B\n1D
3 5 0 6 6 5A\n6C\n6D

How to split comma separated text into columns on pandas dataframe?

I have a dataframe where one of the columns has its items separated with commas. It looks like:
Data
a,b,c
a,c,d
d,e
a,e
a,b,c,d,e
My goal is to create a matrix that has as header all the unique values from column Data, meaning [a,b,c,d,e]. Then as rows a flag indicating if the value is at that particular row.
The matrix should look like this:
Data
a
b
c
d
e
a,b,c
1
1
1
0
0
a,c,d
1
0
1
1
0
d,e
0
0
0
1
1
a,e
1
0
0
0
1
a,b,c,d,e
1
1
1
1
1
To separate column Data what I did is:
df['data'].str.split(',', expand = True)
Then I don't know how to proceed to allocate the flags to each of the columns.

Maybe you can try this without pivot.
Create the dataframe.
import pandas as pd
import io
s = '''Data
a,b,c
a,c,d
d,e
a,e
a,b,c,d,e'''
df = pd.read_csv(io.StringIO(s), sep = "\s+")
We can use pandas.Series.str.split with expand argument equals to True. And value_counts each rows with axis = 1.
Finally fillna with zero and change the data into integer with astype(int).
df["Data"].str.split(pat = ",", expand=True).apply(lambda x : x.value_counts(), axis = 1).fillna(0).astype(int)
#
a b c d e
0 1 1 1 0 0
1 1 0 1 1 0
2 0 0 0 1 1
3 1 0 0 0 1
4 1 1 1 1 1
And then merge it with the original column.
new = df["Data"].str.split(pat = ",", expand=True).apply(lambda x : x.value_counts(), axis = 1).fillna(0).astype(int)
pd.concat([df, new], axis = 1)
#
Data a b c d e
0 a,b,c 1 1 1 0 0
1 a,c,d 1 0 1 1 0
2 d,e 0 0 0 1 1
3 a,e 1 0 0 0 1
4 a,b,c,d,e 1 1 1 1 1

Use the Series.str.get_dummies() method to return the required matrix of 'a', 'b', ... 'e' columns.
df["Data"].str.get_dummies(sep=',')

If you split the strings into lists, then explode them, it makes pivot possible.
(df.assign(data_list=df.Data.str.split(','))
.explode('data_list')
.pivot_table(index='Data',
columns='data_list',
aggfunc=lambda x: 1,
fill_value=0))
Output
data_list a b c d e
Data
a,b,c 1 1 1 0 0
a,b,c,d,e 1 1 1 1 1
a,c,d 1 0 1 1 0
a,e 1 0 0 0 1
d,e 0 0 0 1 1

You could apply a custom count function for each key:
for k in ["a","b","c","d","e"]:
df[k] = df.apply(lambda row: row["Data"].count(k), axis=1)

Creating a column with conditions over multiple rows

I have the next DataFrame:
pd.DataFrame(['a','a','a','a','b','b','b','c','c','a'])
I need to create a column considering the variation on the other column.
Following this result:
Letter
Number
a
1
a
0
a
0
a
0
b
1
b
0
b
0
c
1
c
0
a
1
Every time the letter change, I need to put a 1.

shift
I'm assuming that df is what OP provided
df = pd.DataFrame(['a','a','a','a','b','b','b','c','c','a'])
Then reasigned the first column to a series letter
letter = df.iloc[:, 0]
pd.DataFrame({
'Letter': letter,
'Number': letter.shift().ne(letter).astype(int)
})
Letter Number
0 a 1
1 a 0
2 a 0
3 a 0
4 b 1
5 b 0
6 b 0
7 c 1
8 c 0
9 a 1

How to extract period and variable name from dataframe column strings for multiindex panel data preparation

I'm new to Python and could not find the answer I'm looking for anywhere.
I have a DataFrame that has the following structure:
df = pd.DataFrame(index=list('abc'), data={'A1': range(3), 'A2': range(3),'B1': range(3), 'B2': range(3), 'C1': range(3), 'C2': range(3)})
df
Out[1]:
A1 A2 B1 B2 C1 C2
a 0 0 0 0 0 0
b 1 1 1 1 1 1
c 2 2 2 2 2 2
Where the numbers are periods and he letters are variables. I'd like to transform the columns in a way, that I split the periods and variables into a multiindex. The desired output would look like that
A B C
1 2 1 2 1 2
a 0 0 0 0 0 0
b 1 1 1 1 1 1
c 2 2 2 2 2 2
I've tried the following:
periods = list(range(1, 3))
df.columns = df.columns.str.replace('\d+', '')
df.columns = pd.MultiIndex.from_product([df.columns, periods])
That seams to be multiplying the columns and raising an ValueError: Length mismatch
in my dataframe I have 72 periods and 12 variables.
Thanks in advance for your help!
Edit: I realized that I haven't been precise enough. I have several columns names something like Impressions1, Impressions2...Impressions72 and hhi1, hhi2...hhi72. So df.columns.str[0],df.columns.str[1] does not work for me, as all column names have a different length. I think the solution might contain regex but I can't figure out how to do it. Any ideas?

Use pd.MultiIndex.from_tuples:
df.columns = pd.MultiIndex.from_tuples(list(zip(df.columns.str[0],df.columns.str[1])))
print(df)
A B C
1 2 1 2 1 2
a 0 0 0 0 0 0
b 1 1 1 1 1 1
c 2 2 2 2 2 2
Alternative:
pd.MultiIndex.from_tuples([tuple(name) for name in df.columns])
or
pd.MultiIndex.from_tuples(map(tuple, df.columns))

You can also use, .str.extract and from_frame:
df.columns = pd.MultiIndex.from_frame(df.columns.str.extract('(.)(.)'), names=[None, None])
Output:
A B C
1 2 1 2 1 2
a 0 0 0 0 0 0
b 1 1 1 1 1 1
c 2 2 2 2 2 2

Here is what actually solved my issue:
df.columns = pd.MultiIndex.from_frame(df.columns.str.extract(r'([a-zA-Z]+)([0-9]+)'), names=[None, None])
Thanks #Scott Boston for your inspiration to the solution!

Splitting a concatenated string into seperated columns using pandas

I have a pandas dataframe consiting of one column containing a string seperated by "/" I would like split these seperated strings into new columns denoted by a boolean (if they exist)
d = {'col1': ["A/B/C", "B/C", "D/B/A", "C/B"]}
dataFrame = pd.DataFrame(data=d)
col1
0 A/B/C
1 B/C
2 D/B/A
3 C/B
the result would be as following:
d = {'A': [1, 0, 1, 0], 'B':[1,1,1,1], 'C':[1,1,0,1], 'D':[0,0,1,0]}
dataFrame = pd.DataFrame(data=d)
A B C D
0 1 1 1 0
1 0 1 1 0
2 1 1 0 1
3 0 1 1 0
I have attempted with pandas.Series.str.split and pandas.pivot but nothing quite returns the result I am looking for. Any help or nudges in the right direction, would be highly appreciated!

Use pandas.Series.str.get_dummies
df.col1.str.get_dummies('/')
A B C D
0 1 1 1 0
1 0 1 1 0
2 1 1 0 1
3 0 1 1 0
Setup
d = {'col1': ["A/B/C", "B/C", "D/B/A", "C/B"]}
df = pd.DataFrame(data=d)