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min/max value of a column based on values of another column, grouped by and transformed in pandas

I'd like to know if I can do all this in one line, rather than multiple lines.
my dataframe:
import pandas as pd
df = pd.DataFrame({'ID' : [1,1,1,1,1,1,2,2,2,2,2,2]
,'A': [1, 2, 3, 10, np.nan, 5 , 20, 6, 7, np.nan, np.nan, np.nan]
, 'B': [0,1,1,0,1,1,1,1,1,0,1,0]
, 'desired_output' : [5,5,5,5,5,5,20,20,20,20,20,20]})
df
ID A B desired_output
0 1 1.0 0 5
1 1 2.0 1 5
2 1 3.0 1 5
3 1 10.0 0 5
4 1 NaN 1 5
5 1 5.0 1 5
6 2 20.0 1 20
7 2 6.0 1 20
8 2 7.0 1 20
9 2 NaN 0 20
10 2 NaN 1 20
11 2 NaN 0 20
I'm trying to find the maximum value of column A, for values of column B == 1, group by column ID, and transform the results directly so that the value is back in the dataframe without extra merging et al.
something like the following (but without getting errors!)
df['desired_output'] = df.groupby('ID').A.where(df.B == 1).transform('max') ## this gives error
The max function should ignore the NaNs as well. I wonder if I'm trying too much in one line, but one can hope there is a way for a beautiful code.
EDIT:
I can get a very similar output by changing the where clause:
df['desired_output'] = df.where(df.B == 1).groupby('ID').A.transform('max') ## this works but output is not what i want
but the output is not exactly what I want. desired_output should not have any NaN, unless all values of A are NaN for when B == 1.

Here is a way to do it:
import pandas as pd
import numpy as np
df = pd.DataFrame({
'ID' : [1,1,1,1,1,1,2,2,2,2,2,2],
'A': [1, 2, 3, 10, np.nan, 5 , 20, 6, 7, np.nan, np.nan, np.nan],
'B': [0,1,1,0,1,1,1,1,1,0,1,0],
'desired_output' : [5,5,5,5,5,5,20,20,20,20,20,20]
})
df['output'] = df[df.B == 1].groupby('ID').A.max()[df.ID].array
df
Result:
ID A B desired_output output
0 1 1.0 0 5 5.0
1 1 2.0 1 5 5.0
2 1 3.0 1 5 5.0
3 1 10.0 0 5 5.0
4 1 NaN 1 5 5.0
5 1 5.0 1 5 5.0
6 2 20.0 1 20 20.0
7 2 6.0 1 20 20.0
8 2 7.0 1 20 20.0
9 2 NaN 0 20 20.0
10 2 NaN 1 20 20.0
11 2 NaN 0 20 20.0
Decomposition:
df[df.B == 1] # start by filtering on B
.groupby('ID') # group by ID
.A.max() # get max values in column A
[df.ID] # recast the result on ID series shape
.array # fetch the raw values from the Series
Important note: it relies on the fact that the index is as in the given example, that is, sorted, starting from 0, with a 1 increment. You will have to reset_index() of your DataFrame before this operation when this is not the case.

How to replace NaN with a value which is meant to increase by 10 of previous non Non value?

Suppose I have a DataFrame with some NaNs:
>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df
0 1 2 3
1 4 NaN NaN
2 NaN NaN 9
The result should be like this which is just +10 of the previous NaN value of the column.
0 1 2 3
1 4 12 13
2 14 12 9
Is there any way to do this using any methods?or do I have to iterate each column?

You can use ffill() to fill the NaNs with the previous non-NaN value, and then a simple mask to increment all by 10:
result = df.ffill()
result[df.isna()] += 10
Output
0 1 2
0 1.0 2.0 3.0
1 4.0 12.0 13.0
2 14.0 12.0 9.0

What is the fastest way to calculate a rolling function with a two dimensional window?

I have a pandas dataframe with two dimensions. I want to calculate the rolling standard deviation along axis 1 while also including datapoints in the rows above and below.
So say I have this df:
data = {'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8],
'C': [9, 10, 11, 12]}
df = pd.DataFrame(data)
print(df)
A B C
0 1 5 9
1 2 6 10
2 3 7 11
3 4 8 12
I want a rectangular window 3 rows high and 2 columns across, moving from left to right. So, for example,
std_df.loc[1, 'C']
would be equal to
np.std([1, 5, 9, 2, 6, 10, 3, 7, 11])
But no idea how to achieve this without very slow iteration

Looks like what you want is pd.shift
import pandas as pd
import numpy as np
data = {'A': [1,2,3,4], 'B': [5,6,7,8], 'C': [9,10,11,12]}
df = pd.DataFrame(data)
print(df)
A B C
0 1 5 9
1 2 6 10
2 3 7 11
3 4 8 12
Shifting the dataframe you provided by 1 yields the row above
print(df.shift(1))
A B C
0 NaN NaN NaN
1 1.0 5.0 9.0
2 2.0 6.0 10.0
3 3.0 7.0 11.0
Similarly, shifting the dataframe you provided by -1 yields the row below
print(df.shift(-1))
A B C
0 2.0 6.0 10.0
1 3.0 7.0 11.0
2 4.0 8.0 12.0
3 NaN NaN NaN
so the code below should do what you're looking for (add_prefix prefixes the column names to make them unique)
above_df = df.shift(1).add_prefix('above_')
below_df = df.shift(-1).add_prefix('below_')
lagged = pd.concat([df, above_df, below_df], axis=1)
lagged['std'] = lagged.apply(np.std, axis=1)
print(lagged)
A B C above_A above_B above_C below_A below_B below_C std
0 1 5 9 NaN NaN NaN 2.0 6.0 10.0 3.304038
1 2 6 10 1.0 5.0 9.0 3.0 7.0 11.0 3.366502
2 3 7 11 2.0 6.0 10.0 4.0 8.0 12.0 3.366502
3 4 8 12 3.0 7.0 11.0 NaN NaN NaN 3.304038

How to add an empty row after a definite row in python dataframe?

I'm working with a huge dataframe in python and sometimes I need to add an empty row or several rows in a definite position to dataframe. For this question I create a small dataframe df in order to show, what I want to achieve.
> df = pd.DataFrame(np.random.randint(10, size = (3,3)), columns =
> ['A','B','C'])
> A B C
> 0 4 5 2
> 1 6 7 0
> 2 8 1 9
Let's say I need to add an empty row, if I have a zero-value in the column 'C'. Here the empty row should be added after the second row. So at the end I want to have a new dataframe like:
>new_df
> A B C
> 0 4 5 2
> 1 6 7 0
> 2 nan nan nan
> 3 8 1 9
I tried with concat and append, but I didn't get what I want to. Could you help me please?

You can try in this way:
l = df[df['C']==0].index.tolist()
for c, i in enumerate(l):
dfs = np.split(df, [i+1+c])
df = pd.concat([dfs[0], pd.DataFrame([[np.NaN, np.NaN, np.NaN]], columns=df.columns), dfs[1]], ignore_index=True)
print df
Input:
A B C
0 4 3 0
1 4 0 4
2 4 4 2
3 3 2 1
4 3 1 2
5 4 1 4
6 1 0 4
7 0 2 0
8 2 0 3
9 4 1 3
Output:
A B C
0 4.0 3.0 0.0
1 NaN NaN NaN
2 4.0 0.0 4.0
3 4.0 4.0 2.0
4 3.0 2.0 1.0
5 3.0 1.0 2.0
6 4.0 1.0 4.0
7 1.0 0.0 4.0
8 0.0 2.0 0.0
9 NaN NaN NaN
10 2.0 0.0 3.0
11 4.0 1.0 3.0
Last thing: it can happen that the last row has 0 in 'C', so you can add:
if df["C"].iloc[-1] == 0 :
df.loc[len(df)] = [np.NaN, np.NaN, np.NaN]

Try using slice.
First, you need to find the rows where C == 0. So let's create a bool df for this. I'll just name it 'a':
a = (df['C'] == 0)
So, whenever C == 0, a == True.
Now we need to find the index of each row where C == 0, create an empty row and add it to the df:
df2 = df.copy() #make a copy because we want to be safe here
for i in df.loc[a].index:
empty_row = pd.DataFrame([], index=[i]) #creating the empty data
j = i + 1 #just to get things easier to read
df2 = pd.concat([df2.ix[:i], empty_row, df2.ix[j:]]) #slicing the df
df2 = df2.reset_index(drop=True) #reset the index
I must say... I don't know the size of your df and if this is fast enough, but give it a try

In case you know the index where you want to insert a new row, concat can be a solution.
Example dataframe:
df = pd.DataFrame({'A': [1, 2, 3], 'B': [4, 5, 6], 'C': [7, 8, 9]})
# A B C
# 0 1 4 7
# 1 2 5 8
# 2 3 6 9
Your new row as a dataframe with index 1:
new_row = pd.DataFrame({'A': np.nan, 'B': np.nan,'C': np.nan}, index=[1])
Inserting your new row after the second row:
new_df = pd.concat([df.loc[:1], new_row, df.loc[2:]]).reset_index(drop=True)
# A B C
# 0 1.0 4.0 7.0
# 1 2.0 5.0 8.0
# 2 NaN NaN NaN
# 3 3.0 6.0 9.0

something like this should work for you:
for key, row in df.iterrows():
if row['C'] == 0:
df.loc[key+1] = pd.Series([np.nan])

How to replace NaNs by preceding or next values in pandas DataFrame?

Suppose I have a DataFrame with some NaNs:
>>> import pandas as pd
>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df
0 1 2
0 1 2 3
1 4 NaN NaN
2 NaN NaN 9
What I need to do is replace every NaN with the first non-NaN value in the same column above it. It is assumed that the first row will never contain a NaN. So for the previous example the result would be
0 1 2
0 1 2 3
1 4 2 3
2 4 2 9
I can just loop through the whole DataFrame column-by-column, element-by-element and set the values directly, but is there an easy (optimally a loop-free) way of achieving this?

You could use the fillna method on the DataFrame and specify the method as ffill (forward fill):
>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df.fillna(method='ffill')
0 1 2
0 1 2 3
1 4 2 3
2 4 2 9
This method...
propagate[s] last valid observation forward to next valid
To go the opposite way, there's also a bfill method.
This method doesn't modify the DataFrame inplace - you'll need to rebind the returned DataFrame to a variable or else specify inplace=True:
df.fillna(method='ffill', inplace=True)

The accepted answer is perfect. I had a related but slightly different situation where I had to fill in forward but only within groups. In case someone has the same need, know that fillna works on a DataFrameGroupBy object.
>>> example = pd.DataFrame({'number':[0,1,2,nan,4,nan,6,7,8,9],'name':list('aaabbbcccc')})
>>> example
name number
0 a 0.0
1 a 1.0
2 a 2.0
3 b NaN
4 b 4.0
5 b NaN
6 c 6.0
7 c 7.0
8 c 8.0
9 c 9.0
>>> example.groupby('name')['number'].fillna(method='ffill') # fill in row 5 but not row 3
0 0.0
1 1.0
2 2.0
3 NaN
4 4.0
5 4.0
6 6.0
7 7.0
8 8.0
9 9.0
Name: number, dtype: float64

You can use pandas.DataFrame.fillna with the method='ffill' option. 'ffill' stands for 'forward fill' and will propagate last valid observation forward. The alternative is 'bfill' which works the same way, but backwards.
import pandas as pd
df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
df = df.fillna(method='ffill')
print(df)
# 0 1 2
#0 1 2 3
#1 4 2 3
#2 4 2 9
There is also a direct synonym function for this, pandas.DataFrame.ffill, to make things simpler.

One thing that I noticed when trying this solution is that if you have N/A at the start or the end of the array, ffill and bfill don't quite work. You need both.
In [224]: df = pd.DataFrame([None, 1, 2, 3, None, 4, 5, 6, None])
In [225]: df.ffill()
Out[225]:
0
0 NaN
1 1.0
...
7 6.0
8 6.0
In [226]: df.bfill()
Out[226]:
0
0 1.0
1 1.0
...
7 6.0
8 NaN
In [227]: df.bfill().ffill()
Out[227]:
0
0 1.0
1 1.0
...
7 6.0
8 6.0

Only one column version
Fill NAN with last valid value
df[column_name].fillna(method='ffill', inplace=True)
Fill NAN with next valid value
df[column_name].fillna(method='backfill', inplace=True)

Just agreeing with ffill method, but one extra info is that you can limit the forward fill with keyword argument limit.
>>> import pandas as pd
>>> df = pd.DataFrame([[1, 2, 3], [None, None, 6], [None, None, 9]])
>>> df
0 1 2
0 1.0 2.0 3
1 NaN NaN 6
2 NaN NaN 9
>>> df[1].fillna(method='ffill', inplace=True)
>>> df
0 1 2
0 1.0 2.0 3
1 NaN 2.0 6
2 NaN 2.0 9
Now with limit keyword argument
>>> df[0].fillna(method='ffill', limit=1, inplace=True)
>>> df
0 1 2
0 1.0 2.0 3
1 1.0 2.0 6
2 NaN 2.0 9

ffill now has it's own method pd.DataFrame.ffill
df.ffill()
0 1 2
0 1.0 2.0 3.0
1 4.0 2.0 3.0
2 4.0 2.0 9.0

You can use fillna to remove or replace NaN values.
NaN Remove
import pandas as pd
df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
df.fillna(method='ffill')
0 1 2
0 1.0 2.0 3.0
1 4.0 2.0 3.0
2 4.0 2.0 9.0
NaN Replace
df.fillna(0) # 0 means What Value you want to replace
0 1 2
0 1.0 2.0 3.0
1 4.0 0.0 0.0
2 0.0 0.0 9.0
Reference pandas.DataFrame.fillna

There's also pandas.Interpolate, which I think gives one more control
import pandas as pd
df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
df=df.interpolate(method="pad",limit=None, downcast="infer") #downcast keeps dtype as int
print(df)
0 1 2
0 1 2 3
1 4 2 3
2 4 2 9

In my case, we have time series from different devices but some devices could not send any value during some period. So we should create NA values for every device and time period and after that do fillna.
df = pd.DataFrame([["device1", 1, 'first val of device1'], ["device2", 2, 'first val of device2'], ["device3", 3, 'first val of device3']])
df.pivot(index=1, columns=0, values=2).fillna(method='ffill').unstack().reset_index(name='value')
Result:
0 1 value
0 device1 1 first val of device1
1 device1 2 first val of device1
2 device1 3 first val of device1
3 device2 1 None
4 device2 2 first val of device2
5 device2 3 first val of device2
6 device3 1 None
7 device3 2 None
8 device3 3 first val of device3