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Pandas: Fillna with local average if a condition is met

Let's say I have data like this:
df = pd.DataFrame({'col1': [5, np.nan, 2, 2, 5, np.nan, 4], 'col2':[1,3,np.nan,np.nan,5,np.nan,4]})
print(df)
col1 col2
0 5.0 1.0
1 NaN 3.0
2 2.0 NaN
3 2.0 NaN
4 5.0 5.0
5 NaN NaN
6 4.0 4.0
How can I use fillna() to replace NaN values with the average of the prior and the succeeding value if both of them are not NaN ?
The result would look like this:
col1 col2
0 5.0 1.0
1 3.5 3.0
2 2.0 NaN
3 2.0 NaN
4 5.0 5.0
5 4.5 4.5
6 4.0 4.0
Also, is there a way of calculating the average from the previous n and succeeding n values (if they are all not NaN) ?

We can shift the dataframe forward and backwards. Then add these together and divide them by two and use that to fillna:
s1, s2 = df.shift(), df.shift(-1)
df = df.fillna((s1 + s2) / 2)
col1 col2
0 5.0 1.0
1 3.5 3.0
2 2.0 NaN
3 2.0 NaN
4 5.0 5.0
5 4.5 4.5
6 4.0 4.0

Conditional pairwise calculations in pandas

For example, I have 2 dfs:
df1
ID,col1,col2
1,5,9
2,6,3
3,7,2
4,8,5
and another df is
df2
ID,col1,col2
1,11,9
2,12,7
3,13,2
I want to calculate first pairwise subtraction from df2 to df1. I am using scipy.spatial.distance using a function subtract_
def subtract_(a, b):
return abs(a - b)
d1_s = df1[['col1']]
d2_s = df2[['col1']]
dist = cdist(d1_s, d2_s, metric=subtract_)
dist_df = pd.DataFrame(dist, columns= d2_s.values.ravel())
print(dist_df)
11 12 13
6.0 7.0 8.0
5.0 6.0 7.0
4.0 5.0 6.0
3.0 4.0 5.0
Now, I want to check, these new columns name like 11,12 and 13. I am checking if there is any values in this new dataframe less than 5. If there is, then I want to do further calculations. Like this.
For example, here for columns name '11', less than 5 value is 4 which is at rows 3. Now in this case, I want to subtract columns name ('col2') of df1 but at row 3, in this case it would be value 2. I want to subtract this value 2 with df2(col2) but at row 1 (because column name '11') was from value at row 1 in df2.
My for loop is so complex for this. It would be great, if there would be some easier way in pandas.
Any help, suggestions would be great.
The expected new dataframe is this
0,1,2
Nan,Nan,Nan
Nan,Nan,Nan
(2-9)=-7,Nan,Nan
(5-9)=-4,(5-7)=-2,Nan

Similar to Ben's answer, but with np.where:
pd.DataFrame(np.where(dist_df<5, df1.col2.values[:,None] - df2.col2.values, np.nan),
index=dist_df.index,
columns=dist_df.columns)
Output:
11 12 13
0 NaN NaN NaN
1 NaN NaN NaN
2 -7.0 NaN NaN
3 -4.0 -2.0 NaN

In your case using numpy with mask
df.mask(df<5,df-(df1.col2.values[:,None]+df2.col2.values))
Out[115]:
11 12 13
0 6.0 7.0 8.0
1 5.0 6.0 7.0
2 -7.0 5.0 6.0
3 -11.0 -8.0 5.0
Update
Newdf=(df-(-df1.col2.values[:,None]+df2.col2.values)-df).where(df<5)
Out[148]:
11 12 13
0 NaN NaN NaN
1 NaN NaN NaN
2 -7.0 NaN NaN
3 -4.0 -2.0 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

How do I merge multiple pandas dataframe columns

I have a dataframe similar to the one seen below.
In[2]: df = pd.DataFrame({'P1': [1, 2, None, None, None, None],'P2': [None, None, 3, 4, None, None],'P3': [None, None, None, None, 5, 6]})
Out[2]:
P1 P2 P3
0 1.0 NaN NaN
1 2.0 NaN NaN
2 NaN 3.0 NaN
3 NaN 4.0 NaN
4 NaN NaN 5.0
5 NaN NaN 6.0
And I am trying to merge all of the columns into a single P column in a new dataframe (see below).
P
0 1.0
1 2.0
2 3.0
3 4.0
4 5.0
5 6.0
In my actual code, I have an arbitrary list of columns that should be merged, not necessarily P1, P2, and P3 (between 1 and 5 columns). I've tried something along the following lines:
new_series = pd.Series()
desired_columns = ['P1', 'P2', 'P3']
for col in desired_columns:
other_series=df[col]
new_series = new_series.align(other_series)
However this results in a tuple of Series objects, and neither of them appear to contain the data I need. I could iterate through every row, then check each column, but I feel that there is likely an easy pandas solution that I am missing.

If there is only one non None value per row forward filling Nones and select last column by position:
df['P'] = df[['P1', 'P2', 'P3']].ffill(axis=1).iloc[:, -1]
print (df)
P1 P2 P3 P
0 1.0 NaN NaN 1.0
1 2.0 NaN NaN 2.0
2 NaN 3.0 NaN 3.0
3 NaN 4.0 NaN 4.0
4 NaN NaN 5.0 5.0
5 NaN NaN 6.0 6.0

Another alternate solution:
So, if we are not column specific within the DataFrame to choose about then we can use bfill() function to populate the non-nan values in the dataframe across columns So, when axis='columns', then the current nan cells will be filled from the value present in the next column in the same row.
>>> df['P'] = df.bfill(axis=1).iloc[:, 0]
>>> df
P1 P2 P3 P
0 1.0 NaN NaN 1.0
1 2.0 NaN NaN 2.0
2 NaN 3.0 NaN 3.0
3 NaN 4.0 NaN 4.0
4 NaN NaN 5.0 5.0
5 NaN NaN 6.0 6.0

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