How to merge overlapping columns - python

I have two datasets like this
import pandas as pd
import numpy as np
df1 = pd.DataFrame({'id': [1, 2,3,4,5], 'first': [np.nan,np.nan,1,0,np.nan], 'second': [1,np.nan,np.nan,np.nan,0]})
df2 = pd.DataFrame({'id': [1, 2,3,4,5, 6], 'first': [np.nan,1,np.nan,np.nan,0, 1], 'third': [1,0,np.nan,1,1, 0]})
And I want to get
result = pd.merge(df1, df2, left_index=True, right_index=True,on='id', how= 'outer')
result['first']= result[["first_x", "first_y"]].sum(axis=1)
result.loc[(result['first_x'].isnull()) & (result['first_y'].isnull()), 'first'] = np.nan
result.drop(['first_x','first_y'] , 1)
id second third first
0 1 1.0 1.0 NaN
1 2 NaN 0.0 1.0
2 3 NaN NaN 1.0
3 4 NaN 1.0 0.0
4 5 0.0 1.0 0.0
5 6 NaN 0.0 1.0
The problem is that the real dataset includes about 200 variables and my way is very long. How to make it easier? Thanks


You should be able to use combine_first:
>>> df1.set_index('id').combine_first(df2.set_index('id'))
first second third
id
1 NaN 1 1
2 1 NaN 0
3 1 NaN NaN
4 0 NaN 1
5 0 0 1
6 1 NaN 0


Should probably use combine_first as mentioned by Alexander. If you want to keep id as a column, you would just use:
merged = df1.merge(df2)

Related

Python Pandas: divide a Series by a Dataframe

I have a Series and a Dataframe that share the same index:
s = pd.Series([300, 300])
df = pd.DataFrame({
'A': [10,20],
'B': [20,30]
})
When I do s.div(df), I see:
A B 0 1
0 NaN NaN NaN NaN
1 NaN NaN NaN NaN
I expect:
A B
0 30 15
1 15 10
pandas.__version__: 1.3.4.

Use DataFrame.rdiv for divide from right side:
df1 = df.rdiv(s, axis=0)
print (df1)
A B
0 30.0 15.0
1 15.0 10.0

Duplicate Row Value to Next Null Rows [duplicate]

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

Pandas Replacing Current Value with Previous Value if NaN [duplicate]

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

Delete values from pandas dataframe based on logical operation

I want to delete the values that are greater than a certain threshold from a pandas dataframe. Is there an efficient way to perform this? I am doing it with apply and lambda, which works fine but a bit slow for a large dataframe and I feel like there must be a better method.
df = pd.DataFrame({'A': [1,2,3,4,5], 'B': [1,2,3,4,5]})
df
A B
0 1 1
1 2 2
2 3 3
3 4 4
4 5 5
How can this be done without apply and lambda?
df['A'] = df.apply(lambda x: x['A'] if x['A'] < 3 else None, axis=1)
df
A B
0 1.0 1
1 2.0 2
2 NaN 3
3 NaN 4
4 NaN 5

Use a boolean mask against the df:
In[21]:
df[df<3]
Out[21]:
A
0 1.0
1 2.0
2 NaN
3 NaN
4 NaN
Here where the boolean condition is not met a False is returned, this will just mask out the df value returning NaN
If you actually want to drop these rows then self-assign:
df = df[df<3]
To compare a specific column:
In[22]:
df[df['A']<3]
Out[22]:
A
0 1
1 2
If you want NaN in the removed rows then you can use a trick where a double square brackets will return a single column df so we can mask the df:
In[25]:
df[df[['A']]<3]
Out[25]:
A
0 1.0
1 2.0
2 NaN
3 NaN
4 NaN
If you have multiple columns then the above won't work as the boolean mask has to match the orig df, in which case you can reindex against the orig df index:
In[31]:
df = pd.DataFrame({'A': [1,2,3,4,5], 'B': [1,2,3,4,5]})
df[df['A']<3].reindex(df.index)
Out[31]:
A B
0 1.0 1.0
1 2.0 2.0
2 NaN NaN
3 NaN NaN
4 NaN NaN
EDIT
You've updated your question again, if you want to just overwrite the single column:
In[32]:
df = pd.DataFrame({'A': [1,2,3,4,5], 'B': [1,2,3,4,5]})
df['A'] = df.loc[df['A'] < 3,'A']
df
Out[32]:
A B
0 1.0 1
1 2.0 2
2 NaN 3
3 NaN 4
4 NaN 5

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

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