Develop Reference

Pandas transformation of a column to row in a .rolling() fashion

Imagine I have the following Series:
inp = pd.Series(np.arange(10))
What I want to do, is to transform it to a np.array in the following way:
input
0
1
2
3
4
0
NaN
NaN
NaN
Nan
0
1
NaN
NaN
NaN
0
1
2
NaN
NaN
0
1
2
3
NaN
0
1
2
3
4
0
1
2
3
4
5
1
2
3
4
5
6
2
3
4
5
6
...and so forth.
The column called input is not expected in the output, but I placed it here to make my inquiry more clear.
What I tried is the following:
matrix = [x.to_numpy() for x in list(inp.rolling(window=5, min_periods=5))]
Problem is, i can't use np.stack() on matrix, as (even though I passed min_periods=5) the shape of every item in the list is different.
Also I feel like I am overlooking a very simple pandas command :D.
Thank you very much!
EDIT:
My current workaround is a custom function. I guess there are way better solutions than this one:
def rolling_transform_series(x):
length = len(x)
array = []
for idx in range(length):
s = x[idx-5:idx]
if idx < 5:
s = np.r_[np.zeros(5-idx), x[:idx]]
s[s==0] = np.nan
array.append(s)
return np.array(array)
df = inp.apply(rolling_transform_series)

You can try it with shift:
import pandas as pd
inp = pd.Series(np.arange(10))
pd.DataFrame([inp.shift(s) for s in range(4,-6,-1)])
0 1 2 3 4 5 6 7 8 9
0 NaN NaN NaN NaN 0.0 1.0 2.0 3.0 4.0 5.0
1 NaN NaN NaN 0.0 1.0 2.0 3.0 4.0 5.0 6.0
2 NaN NaN 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
3 NaN 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
4 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0
5 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 NaN
6 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 NaN NaN
7 3.0 4.0 5.0 6.0 7.0 8.0 9.0 NaN NaN NaN
8 4.0 5.0 6.0 7.0 8.0 9.0 NaN NaN NaN NaN
9 5.0 6.0 7.0 8.0 9.0 NaN NaN NaN NaN NaN

Manipulating value in a column based on a rule

I have 3 columns -A, B and C in a pandas dataframe. What i want to do is, where ever A is not null AND B|C are not null, that row in A should be set to null.
if(dffinal['A'].loc[dffinal['A'].notnull()] &
(dffinal['B'].loc[dffinal['B'].notnull()] |
dffinal['C'].loc[dffinal['C'].notnull()])):
dffinal['A'] = np.nan
this is the error I'm getting: cannot do a non-empty take from an empty axes.

Use df.loc[]:
df.loc[df.A.notna() & (df.B.notna()|df.C.notna()),'A']=np.nan

Here first condition is not necessary, so solution should be simplify:
dffinal = pd.DataFrame({
'A':[np.nan,np.nan,4,5,5,np.nan],
'B':[7,np.nan,np.nan,4,np.nan,np.nan],
'C':[1,3,5,7,np.nan,np.nan],
})
print (dffinal)
A B C
0 NaN 7.0 1.0
1 NaN NaN 3.0
2 4.0 NaN 5.0
3 5.0 4.0 7.0
4 5.0 NaN NaN
5 NaN NaN NaN
mask = (dffinal['B'].notnull() | dffinal['C'].notnull())
dffinal.loc[mask, 'A'] = np.nan
print (dffinal)
A B C
0 NaN 7.0 1.0
1 NaN NaN 3.0
2 NaN NaN 5.0
3 NaN 4.0 7.0
4 5.0 NaN NaN
5 NaN NaN NaN
Same output like in first condition:
mask = dffinal['A'].notnull() & (dffinal['B'].notnull() | dffinal['C'].notnull())
dffinal.loc[mask, 'A'] = np.nan
print (dffinal)
A B C
0 NaN 7.0 1.0
1 NaN NaN 3.0
2 NaN NaN 5.0
3 NaN 4.0 7.0
4 5.0 NaN NaN
5 NaN NaN NaN

Pandas combine two columns

I have following database:
df = pandas.DataFrame({'Buy':[10,np.nan,2,np.nan,np.nan,4],'Sell':[np.nan,7,np.nan,9,np.nan,np.nan]})
Out[37]:
Buy Sell
0 10.0 NaN
1 NaN 7.0
2 2.0 NaN
3 NaN 9.0
4 NaN NaN
5 4.0 NaN
I want o create two more columns called Quant and B/S
for Quant it is working fine as follows:
df['Quant'] = df['Buy'].fillna(df['Sell']) # Fetch available value from both column and if both values are Nan then output is Nan.
Output is:
df
Out[39]:
Buy Sell Quant
0 10.0 NaN 10.0
1 NaN 7.0 7.0
2 2.0 NaN 2.0
3 NaN 9.0 9.0
4 NaN NaN NaN
5 4.0 NaN 4.0
But I want to create B/S on the basis of "from which column they have taken value while creating Quant"

You can perform an equality test and feed into numpy.where:
df['B/S'] = np.where(df['Quant'] == df['Buy'], 'B', 'S')
For the case where both values are null, you can use an additional step:
df.loc[df[['Buy', 'Sell']].isnull().all(1), 'B/S'] = np.nan
Example
from io import StringIO
import pandas as pd
mystr = StringIO("""Buy Sell
10 nan
nan 8
4 nan
nan 5
nan 7
3 nan
2 nan
nan nan""")
df = pd.read_csv(mystr, delim_whitespace=True)
df['Quant'] = df['Buy'].fillna(df['Sell'])
df['B/S'] = np.where(df['Quant'] == df['Buy'], 'B', 'S')
df.loc[df[['Buy', 'Sell']].isnull().all(1), 'B/S'] = np.nan
Result
print(df)
Buy Sell Quant B/S
0 10.0 NaN 10.0 B
1 NaN 8.0 8.0 S
2 4.0 NaN 4.0 B
3 NaN 5.0 5.0 S
4 NaN 7.0 7.0 S
5 3.0 NaN 3.0 B
6 2.0 NaN 2.0 B
7 NaN NaN NaN NaN

replace nan in pandas dataframe

given the dataframe df
df = pd.DataFrame(data=[[np.nan,1],
[np.nan,np.nan],
[1,2],
[2,3],
[np.nan,np.nan],
[np.nan,np.nan],
[3,4],
[4,5],
[np.nan,np.nan],
[np.nan,np.nan]],columns=['A','B'])
df
Out[16]:
A B
0 NaN 1.0
1 NaN NaN
2 1.0 2.0
3 2.0 3.0
4 NaN NaN
5 NaN NaN
6 3.0 4.0
7 4.0 5.0
8 NaN NaN
9 NaN NaN
I would need to replace the nan using the following rules:
1) if nan is at the beginning replace with the first values after the nan
2) if nan is in the middle of 2 or more values replace the nan with the average of these values
3) if nan is at the end replace with the last value
df
Out[16]:
A B
0 1.0 1.0
1 1.0 1.5
2 1.0 2.0
3 2.0 3.0
4 2.5 3.5
5 2.5 3.5
6 3.0 4.0
7 4.0 5.0
8 4.0 5.0
9 4.0 5.0

Use add between forward filling and backfilling values, then divide by 2 and last replace last and first NaNs:
df = df.bfill().add(df.ffill()).div(2).ffill().bfill()
print (df)
A B
0 1.0 1.0
1 1.0 1.5
2 1.0 2.0
3 2.0 3.0
4 2.5 3.5
5 2.5 3.5
6 3.0 4.0
7 4.0 5.0
8 4.0 5.0
9 4.0 5.0
Detail:
print (df.bfill().add(df.ffill()))
A B
0 NaN 2.0
1 NaN 3.0
2 2.0 4.0
3 4.0 6.0
4 5.0 7.0
5 5.0 7.0
6 6.0 8.0
7 8.0 10.0
8 NaN NaN
9 NaN NaN

pandas backfill NaN by incrementing the last value

I have a data frame:
A B C
Timestamp
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN 5
4 NaN NaN 4
5 NaN 3 3
6 NaN 2 NaN
7 3 1 NaN
8 2 NaN NaN
9 1 NaN NaN
I would like to backfill it by incrementing the last available value in each column so it looks like this:
A B C
Timestamp
1 9 7 7
2 8 6 6
3 7 5 5
4 6 4 4
5 5 3 3
6 4 2 NaN
7 3 1 NaN
8 2 NaN NaN
9 1 NaN NaN

Let's try this:
df1 = df1[::-1].fillna(method='ffill')
(df1 + (df1 == df1.shift()).cumsum()).sort_index()
Output:
A B C
Timestamp
1 9.0 7.0 7.0
2 8.0 6.0 6.0
3 7.0 5.0 5.0
4 6.0 4.0 4.0
5 5.0 3.0 3.0
6 4.0 2.0 NaN
7 3.0 1.0 NaN
8 2.0 NaN NaN
9 1.0 NaN NaN

You can try this:
def bfill_increment(col):
col_null = col.isnull()[::-1]
groups = col_null.diff().fillna(0).cumsum()
return col_null.groupby(groups).cumsum()[::-1] + col.bfill()
df.apply(bfill_increment)