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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.

pandas filling nan with previous row value multiplied with another column

I have dataframe for which I want to fill nan with values from previous rows mulitplied with pct_change column
col_to_fill pct_change
0 1 NaN
1 2 1.0
2 10 0.5
3 nan 0.5
4 nan 1.3
5 nan 2
6 5 3
so for 3rd row 10*0.5 = 5 and use that filled value to fill next rows if its nan.
col_to_fill pct_change
0 1 NaN
1 2 1.0
2 10 0.5
3 5 0.5
4 6.5 1.3
5 13 2
6 5 3
I have used this
while df['col_to_fill'].isna().sum() > 0:
df.loc[df['col_to_fill'].isna(), 'col_to_fill'] = df['col_to_fill'].shift(1) * df['pct_change']
but Its taking too much time as its only filling those row whos previous row are nonnan in one loop.

Try with cumprod after ffill
s = df.col_to_fill.ffill()*df.loc[df.col_to_fill.isna(),'pct_change'].cumprod()
df.col_to_fill.fillna(s, inplace=True)
df
Out[90]:
col_to_fill pct_change
0 1.0 NaN
1 2.0 1.0
2 10.0 0.5
3 5.0 0.5
4 6.5 1.3
5 13.0 2.0
6 5.0 3.0

Perform arithmetic operations on null values

When i am trying to do arithmetic operation including two or more columns facing problem with null values.
One more thing which i want to mention here that i don't want to fill missed/null values.
Actually i want something like 1 + np.nan = 1 but it is giving np.nan. I tried to solve it by np.nansum but it didn't work.
df = pd.DataFrame({"a":[1,2,3,4],"b":[1,2,np.nan,np.nan]})
df
Out[6]:
a b c
0 1 1.0 2.0
1 2 2.0 4.0
2 3 NaN NaN
3 4 NaN NaN
And,
df["d"] = np.nansum([df.a + df.b])
df
Out[13]:
a b d
0 1 1.0 6.0
1 2 2.0 6.0
2 3 NaN 6.0
3 4 NaN 6.0
But i want actually like,
df
Out[10]:
a b c
0 1 1.0 2.0
1 2 2.0 4.0
2 3 NaN 3.0
3 4 NaN 4.0

The np.nansum here calculated the sum, of the entire column. You do not want that, you probably want to call the np.nansum on the two columns, like:
df['d'] = np.nansum((df.a, df.b), axis=0)
This then yield the expected:
>>> df
a b d
0 1 1.0 2.0
1 2 2.0 4.0
2 3 NaN 3.0
3 4 NaN 4.0

Simply use DataFrame.sum over axis=1:
df['c'] = df.sum(axis=1)
Output
a b c
0 1 1.0 2.0
1 2 2.0 4.0
2 3 NaN 3.0
3 4 NaN 4.0

If dataframe length exceeds threshold, make a new row python

I have a data frame with columns a,b,c,d
a b c d
1 2 nan nan
2 3 4 5
4 5 nan nan
how do i reshape into 2 columns, when i am not aware of the number of rows that the result will give. (big data)
output:
a b
1 2
2 3
4 5
4 5

Numpy's reshape
pd.DataFrame(df.values.reshape(-1, 2), columns=['a', 'b']).dropna()
a b
0 1.0 2.0
2 2.0 3.0
3 4.0 5.0
4 4.0 5.0

pandas rolling apply to allow nan

I have a very simple Pandas Series:
xx = pd.Series([1, 2, np.nan, np.nan, 3, 4, 5])
If I run this I get what I want:
>>> xx.rolling(3,1).mean()
0 1.0
1 1.5
2 1.5
3 2.0
4 3.0
5 3.5
6 4.0
But if I have to use .apply() I cannot get it to work by ignoring NaNs in the mean() operation:
>>> xx.rolling(3,1).apply(np.mean)
0 1.0
1 1.5
2 NaN
3 NaN
4 NaN
5 NaN
6 4.0
>>> xx.rolling(3,1).apply(lambda x : np.mean(x))
0 1.0
1 1.5
2 NaN
3 NaN
4 NaN
5 NaN
6 4.0
What should I do in order to both use .apply() and have the result in the first output? My actual problem is more complicated that I have to use .apply() to realize but it boils down to this issue.

You can use np.nanmean()
xx.rolling(3,1).apply(lambda x : np.nanmean(x))
Out[59]:
0 1.0
1 1.5
2 1.5
3 2.0
4 3.0
5 3.5
6 4.0
dtype: float64
If you have to process the nans explicitly, you can do:
xx.rolling(3,1).apply(lambda x : np.mean(x[~np.isnan(x)]))
Out[94]:
0 1.0
1 1.5
2 1.5
3 2.0
4 3.0
5 3.5
6 4.0
dtype: float64