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Concatenate columns skipping pasted rows and columns

I expect to describe well want I need. I have a data frame with the same columns name and another column that works as an index. The data frame looks as follows:
df = pd.DataFrame({'ID':[1,1,1,1,1,2,2,2,3,3,3,3],'X':[1,2,3,4,5,2,3,4,1,3,4,5],'Y':[1,2,3,4,5,2,3,4,5,4,3,2]})
df
Out[21]:
ID X Y
0 1 1 1
1 1 2 2
2 1 3 3
3 1 4 4
4 1 5 5
5 2 2 2
6 2 3 3
7 2 4 4
8 3 1 5
9 3 3 4
10 3 4 3
11 3 5 2
My intention is to copy X as an index or one column (it doesn't matter) and append Y columns from each 'ID' in the following way:

You can try
out = pd.concat([group.rename(columns={'Y': f'Y{name}'}) for name, group in df.groupby('ID')])
out.columns = out.columns.str.replace(r'\d+$', '', regex=True)
print(out)
ID X Y Y Y
0 1 1 1.0 NaN NaN
1 1 2 2.0 NaN NaN
2 1 3 3.0 NaN NaN
3 1 4 4.0 NaN NaN
4 1 5 5.0 NaN NaN
5 2 2 NaN 2.0 NaN
6 2 3 NaN 3.0 NaN
7 2 4 NaN 4.0 NaN
8 3 1 NaN NaN 5.0
9 3 3 NaN NaN 4.0
10 3 4 NaN NaN 3.0
11 3 5 NaN NaN 2.0

Here's another way to do it:
df_org = pd.DataFrame({'ID':[1,1,1,1,1,2,2,2,3,3,3,3],'X':[1,2,3,4,5,2,3,4,1,3,4,5]})
df = df_org.copy()
for i in set(df_org['ID']):
df1 = df_org[df_org['ID']==i]
col = 'Y'+str(i)
df1.columns = ['ID', col]
df = pd.concat([ df, df1[[col]] ], axis=1)
df.columns = df.columns.str.replace(r'\d+$', '', regex=True)
print(df)
Output:
ID X Y Y Y
0 1 1 1.0 NaN NaN
1 1 2 2.0 NaN NaN
2 1 3 3.0 NaN NaN
3 1 4 4.0 NaN NaN
4 1 5 5.0 NaN NaN
5 2 2 NaN 2.0 NaN
6 2 3 NaN 3.0 NaN
7 2 4 NaN 4.0 NaN
8 3 1 NaN NaN 1.0
9 3 3 NaN NaN 3.0
10 3 4 NaN NaN 4.0
11 3 5 NaN NaN 5.0

Another solution could be as follow.
Get unique values for column ID (stored in array s).
Use np.transpose to repeat column ID n times (n == len(s)) and evaluate the array's matches with s.
Use np.where to replace True with values from df.Y and False with NaN.
Finally, drop the orignal df.Y and rename the new columns as required.
import pandas as pd
import numpy as np
df = pd.DataFrame({'ID':[1,1,1,1,1,2,2,2,3,3,3,3],
'X':[1,2,3,4,5,2,3,4,1,3,4,5],
'Y':[1,2,3,4,5,2,3,4,5,4,3,2]})
s = df.ID.unique()
df[s] = np.where((np.transpose([df.ID]*len(s))==s),
np.transpose([df.Y]*len(s)),
np.nan)
df.drop('Y', axis=1, inplace=True)
df.rename(columns={k:'Y' for k in s}, inplace=True)
print(df)
ID X Y Y Y
0 1 1 1.0 NaN NaN
1 1 2 2.0 NaN NaN
2 1 3 3.0 NaN NaN
3 1 4 4.0 NaN NaN
4 1 5 5.0 NaN NaN
5 2 2 NaN 2.0 NaN
6 2 3 NaN 3.0 NaN
7 2 4 NaN 4.0 NaN
8 3 1 NaN NaN 5.0
9 3 3 NaN NaN 4.0
10 3 4 NaN NaN 3.0
11 3 5 NaN NaN 2.0
If performance is an issue, this method should be faster than this answer, especially when the number of unique values for ID increases.

Create dataframe with hierarchical indices and extra columns from non-hierarchically indexed dataframe

Consider a simple dataframe:
import numpy as np
import pandas as pd
x = pd.DataFrame(np.arange(10).reshape(5,2))
print(x)
0 1
0 0 1
1 2 3
2 4 5
3 6 7
4 8 9
I would like to create a hierarchically indexed dataframe of the form:
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN
where the 'a' columns correspond to the original dataframe columns and the 'b' columns are blank (or nan).
I can certainly create a hierarchically indexed dataframe with all NaNs and loop over the columns of the original dataframe, writing them into
the new dataframe. Is there something more compact than that?

you can do with MultiIndex.from_product
extra_level = ['a', 'b']
new_cols = pd.MultiIndex.from_product([x.columns, extra_level])
x.columns = new_cols[::len(x.columns)] # select all the first element of extra_level
x = x.reindex(columns=new_cols)
print(x)
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN

Very much like #Ben.T I am using MultiIndex.from_product:
x.assign(l='a')
.set_index('l', append=True)
.unstack()
.reindex(pd.MultiIndex.from_product([x.columns.tolist(), ['a','b']]), axis=1)
Output:
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN

How Can I combine two columns is one dataframe?

I have a dataset like this.
A B C A2
1 2 3 4
5 6 7 8
and I want to combine A and A2.
A B C
1 2 3
5 6 7
4
8
how can I combine two columns?
Hope for help. Thank you.

I don't think it is possible directly. But you can do it with a few lines of code:
df = pd.DataFrame({'A':[1,5],'B':[2,6],'C':[3,7],'A2':[4,8]})
df_A2 = df[['A2']]
df_A2.columns = ['A']
df = pd.concat([df.drop(['A2'],axis=1),df_A2])
You will get this if you print df:
A B C
0 1 2.0 3.0
1 5 6.0 7.0
0 4 NaN NaN
1 8 NaN NaN

You could append the last columns after renaming it:
df.append(df[['A2']].set_axis(['A'], axis=1)).drop(columns='A2')
it gives as expected:
A B C
0 1 2.0 3.0
1 5 6.0 7.0
0 4 NaN NaN
1 8 NaN NaN

if the index is not important to you:
import pandas as pd
pd.concat([df[['A','B','C']], df[['A2']].rename(columns={'A2': 'A'})]).reset_index(drop=True)

Appending Pandas DataFrame in a loop

Let's say I have a df such as this:
df = pd.DataFrame({'A': [1,2,3,4,5], 'A_z': [2,3,4,5,6], 'B': [3,4,5,6,7], 'B_z': [4,5,6,7,8],
'C': [5,6,7,8,9], 'C_z': [6,7,8,9,10]})
Which looks like this:
A A_z B B_z C C_z
0 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
What I'm looking to do is create a new df and for each letter (A,B,C) append this new df vertically with the data from the two columns per letter so that it looks like this:
Letter Letter_z
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 3 4
6 4 5
7 5 6
8 6 7
9 7 8
10 5 6
11 6 7
12 7 8
13 8 9
14 9 10
As far as I'm concerned something like this should work fine:
for col in df.columns:
if col[-1] != 'z':
new_df = new_df.append(df[[col, col + '_z']])
However this results in the following mess:
A A_z B B_z C C_z
0 1.0 2.0 NaN NaN NaN NaN
1 2.0 3.0 NaN NaN NaN NaN
2 3.0 4.0 NaN NaN NaN NaN
3 4.0 5.0 NaN NaN NaN NaN
4 5.0 6.0 NaN NaN NaN NaN
0 NaN NaN 3.0 4.0 NaN NaN
1 NaN NaN 4.0 5.0 NaN NaN
2 NaN NaN 5.0 6.0 NaN NaN
3 NaN NaN 6.0 7.0 NaN NaN
4 NaN NaN 7.0 8.0 NaN NaN
0 NaN NaN NaN NaN 5.0 6.0
1 NaN NaN NaN NaN 6.0 7.0
2 NaN NaN NaN NaN 7.0 8.0
3 NaN NaN NaN NaN 8.0 9.0
4 NaN NaN NaN NaN 9.0 10.0
What am I doing wrong? Any help would be really appreciated, cheers.
EDIT:
After the kind help from jezrael the renaming of the columns in his answer got me thinking about a possible way to do it using my original train of thought.
I can now also achieve the new df I want using the following:
for col in df:
if col[-1] != 'z':
d = df[[col, col + '_z']]
d.columns = ['Letter', 'Letter_z']
new_df = new_df.append(d)
The different columns names were clearly what was causing the problem which is something I wasn't aware of at the time. Hope this helps anyone.

One ide is use Series.str.split with expand=True for MultiIndex, then use rename for avoid NaNs and finally new columns names, reshape by DataFrame.stack, sort for correct order by DataFrame.sort_index and last remove MultiIndex:
df.columns = df.columns.str.split('_', expand=True)
df = df.rename(columns=lambda x:'Letter_z' if x == 'z' else 'Letter', level=1)
df = df.stack(0).sort_index(level=[1,0]).reset_index(drop=True)
print (df)
Letter Letter_z
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 3 4
6 4 5
7 5 6
8 6 7
9 7 8
10 5 6
11 6 7
12 7 8
13 8 9
14 9 10
Or if possible simplify problem with reshape all non z values to one column and all z values to another use numpy.ravel:
m = df.columns.str.endswith('_z')
a = df.loc[:, ~m].to_numpy().T.ravel()
b = df.loc[:, m].to_numpy().T.ravel()
df = pd.DataFrame({'Letter': a,'Letter_z': b})
print (df)
Letter Letter_z
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 3 4
6 4 5
7 5 6
8 6 7
9 7 8
10 5 6
11 6 7
12 7 8
13 8 9
14 9 10

You can use the function concat and a list comprehension:
cols = df.columns[~df.columns.str.endswith('_z')]
func = lambda x: 'letter_z' if x.endswith('_z') else 'letter'
pd.concat([df.filter(like=i).rename(func, axis=1) for i in cols])
or
cols = df.columns[~df.columns.str.endswith('_z')]
pd.concat([df.filter(like=i).set_axis(['letter', 'letter_z'], axis=1, inplace=False) for i in cols])

Add Uneven Sized Data Columns in Pandas

I want to add a list as a column to the df dataframe. The list has a different size than the column length.
df =
A B C
1 2 3
5 6 9
4
6 6
8 4
2 3
4
6 6
8 4
D = [11,17,18]
I want the following output
df =
A B C D
1 2 3 11
5 6 9 17
4 18
6 6
8 4
2 3
4
6 6
8 4
I am doing the following to extend the list to the size of the dataframe by adding "nan"
# number of nan value require for the list to match the size of the column
extend_length = df.shape[0]-len(D)
# extend the list
D.extend(extend_length * ['nan'])
# add to the dataframe
df["D"] = D
A B C D
1 2 3 11
5 6 9 17
4 18
6 6 nan
8 4 nan
2 3 nan
4 nan
6 6 nan
8 4 nan
Where "nan" is treated like string but I want it to be empty ot "nan", thus, if I search for number of valid cell in D column it will provide output of 3.

Adding the list as a Series will handle this directly.
D = [11,17,18]
df.loc[:, 'D'] = pd.Series(D)

A simple pd.concat on df and series of D as follows:
pd.concat([df, pd.Series(D, name='D')], axis=1)
or
df.assign(D=pd.Series(D))
Out[654]:
A B C D
0 1 2.0 3.0 11.0
1 5 6.0 9.0 17.0
2 4 NaN NaN 18.0
3 6 NaN 6.0 NaN
4 8 NaN 4.0 NaN
5 2 NaN 3.0 NaN
6 4 NaN NaN NaN
7 6 NaN 6.0 NaN
8 8 NaN 4.0 NaN