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Add all column values repeated of one data frame to other in pandas

Having two data frames:
df1 = pd.DataFrame({'a':[1,2,3],'b':[4,5,6]})
a b
0 1 4
1 2 5
2 3 6
df2 = pd.DataFrame({'c':[7],'d':[8]})
c d
0 7 8
The goal is to add all df2 column values to df1, repeated and create the following result. It is assumed that both data frames do not share any column names.
a b c d
0 1 4 7 8
1 2 5 7 8
2 3 6 7 8

If there are strings columns names is possible use DataFrame.assign with unpack Series created by selecing first row of df2:
df = df1.assign(**df2.iloc[0])
print (df)
a b c d
0 1 4 7 8
1 2 5 7 8
2 3 6 7 8
Another idea is repeat values by df1.index with DataFrame.reindex and use DataFrame.join (here first index value of df2 is same like first index value of df1.index):
df = df1.join(df2.reindex(df1.index, method='ffill'))
print (df)
a b c d
0 1 4 7 8
1 2 5 7 8
2 3 6 7 8
If no missing values in original df is possible use forward filling missing values in last step, but also are types changed to floats, thanks #Dishin H Goyan:
df = df1.join(df2).ffill()
print (df)
a b c d
0 1 4 7.0 8.0
1 2 5 7.0 8.0
2 3 6 7.0 8.0

Pandas - combine columns and put one after another?

I have the following dataframe:
a1,a2,b1,b2
1,2,3,4
2,3,4,5
3,4,5,6
The desirable output is:
a,b
1,3
2,4
3,5
2,4
3,5
4,6
There is a lot of "a" and "b" named headers in the dataframe, the maximum is a50 and b50. So I am looking for the way to combine them all into just "a" and "b".
I think it's possible to do with concat, but I have no idea how to combine it all, putting all the values under each other. I'll be grateful for any ideas.

You can use pd.wide_to_long:
pd.wide_to_long(df.reset_index(), ['a','b'], 'index', 'No').reset_index()[['a','b']]
Output:
a b
0 1 3
1 2 4
2 3 5
3 2 4
4 3 5
5 4 6

First we read the dataframe:
import pandas as pd
from io import StringIO
s = """a1,a2,b1,b2
1,2,3,4
2,3,4,5
3,4,5,6"""
df = pd.read_csv(StringIO(s), sep=',')
Then we stack the columns, and separate the number of the columns from the letter 'a' or 'b':
stacked = df.stack().rename("val").reset_index(1).reset_index()
cols_numbers = pd.DataFrame(stacked
.level_1
.str.split('(\d)')
.apply(lambda l: l[:2])
.tolist(),
columns=["col", "num"])
x = cols_numbers.join(stacked[['val', 'index']])
print(x)
col num val index
0 a 1 1 0
1 a 2 2 0
2 b 1 3 0
3 b 2 4 0
4 a 1 2 1
5 a 2 3 1
6 b 1 4 1
7 b 2 5 1
8 a 1 3 2
9 a 2 4 2
10 b 1 5 2
11 b 2 6 2
Finally, we group by index and num to get two columns a and b, and we fill the first row of the b column with the second value, to get what was expected:
result = (x
.set_index("col", append=True)
.groupby(["index", "num"])
.val
.apply(lambda g:
g
.unstack()
.fillna(method="bfill")
.head(1))
.reset_index(-1, drop=True))
print(result)
col a b
index num
0 1 1.0 3.0
2 2.0 4.0
1 1 2.0 4.0
2 3.0 5.0
2 1 3.0 5.0
2 4.0 6.0
To get rid of the multiindex at the end: result.reset_index(drop=True)

Python pandas: creating a discrete series from a cumulative

I have a data frame where there are several groups of numeric series where the values are cumulative. Consider the following:
df = pd.DataFrame({'Cat': ['A', 'A','A','A', 'B','B','B','B'], 'Indicator': [1,2,3,4,1,2,3,4], 'Cumulative1': [1,3,6,7,2,4,6,9], 'Cumulative2': [1,3,4,6,1,5,7,12]})
In [74]:df
Out[74]:
Cat Cumulative1 Cumulative2 Indicator
0 A 1 1 1
1 A 3 3 2
2 A 6 4 3
3 A 7 6 4
4 B 2 1 1
5 B 4 5 2
6 B 6 7 3
7 B 9 12 4
I need to create discrete series for Cumulative1 and Cumulative2, with starting point being the earliest entry in 'Indicator'.
my Approach is to use diff()
In[82]: df['Discrete1'] = df.groupby('Cat')['Cumulative1'].diff()
Out[82]: df
Cat Cumulative1 Cumulative2 Indicator Discrete1
0 A 1 1 1 NaN
1 A 3 3 2 2.0
2 A 6 4 3 3.0
3 A 7 6 4 1.0
4 B 2 1 1 NaN
5 B 4 5 2 2.0
6 B 6 7 3 2.0
7 B 9 12 4 3.0
I have 3 questions:
How do I avoid the NaN in an elegant/Pythonic way? The correct values are to be found in the original Cumulative series.
Secondly, how do I elegantly apply this computation to all series, say -
cols = ['Cumulative1', 'Cumulative2']
Thirdly, I have a lot of data that needs this computation -- is this the most efficient way?

You do not want to avoid NaNs, you want to fill them with the start values from the "cumulative" column:
df['Discrete1'] = df['Discrete1'].combine_first(df['Cumulative1'])
To apply the operation to all (or select) columns, broadcast it to all columns of interest:
sources = 'Cumulative1', 'Cumulative2'
targets = ["Discrete" + x[len('Cumulative'):] for x in sources]
df[targets] = df.groupby('Cat')[sources].diff()
You still have to condition the NaNs in a loop:
for s,t in zip(sources, targets):
df[t] = df[t].combine_first(df[s])

Pandas - remove row similar to other row

I need to remove all rows from a pandas.DataFrame, which satisfy an unusual condition.
In case there is an exactly the same row, except for it has Nan value in column "C", I want to remove this row.
Given a table:
A B C D
1 2 NaN 3
1 2 50 3
10 20 NaN 30
5 6 7 8
I need to remove the first row, since it has Nan in column C, but there is absolutely same row (second) with real value in column C.
However, 3rd row must stay, because there're no rows with same A, B and D values as it has.
How do you perform this using pandas? Thank you!

You can achieve in using drop_duplicates.
Initial DataFrame:
df=pd.DataFrame(columns=['a','b','c','d'], data=[[1,2,None,3],[1,2,50,3],[10,20,None,30],[5,6,7,8]])
df
a b c d
0 1 2 NaN 3
1 1 2 50 3
2 10 20 NaN 30
3 5 6 7 8
Then you can sort DataFrame by column C. This will drop NaNs to the bottom of column:
df = df.sort_values(['c'])
df
a b c d
3 5 6 7 8
1 1 2 50 3
0 1 2 NaN 3
2 10 20 NaN 30
And then remove duplicates selecting taken into account columns ignoring C and keeping first catched row:
df1 = df.drop_duplicates(['a','b','d'], keep='first')
a b c d
3 5 6 7 8
1 1 2 50 3
2 10 20 NaN 30
But it will be valid only if NaNs are in column C.

You can try fillna along with drop_duplicates
df.bfill().ffill().drop_duplicates(subset=['A', 'B', 'D'], keep = 'last')
This will handle the scenario such as A, B and D values are same but C has non-NaN values in both the rows.
You get
A B C D
1 1 2 50 3
2 10 20 Nan 30
3 5 6 7 8

This feels right to me
notdups = ~df.duplicated(df.columns.difference(['C']), keep=False)
notnans = df.C.notnull()
df[notdups | notnans]
A B C D
1 1 2 50.0 3
2 10 20 NaN 30
3 5 6 7.0 8

Pandas merge on aggregated columns

Let's say I create a DataFrame:
import pandas as pd
df = pd.DataFrame({"a": [1,2,3,13,15], "b": [4,5,6,6,6], "c": ["wish", "you","were", "here", "here"]})
Like so:
a b c
0 1 4 wish
1 2 5 you
2 3 6 were
3 13 6 here
4 15 6 here
... and then group and aggregate by a couple columns ...
gb = df.groupby(['b','c']).agg({"a": lambda x: x.nunique()})
Yielding the following result:
a
b c
4 wish 1
5 you 1
6 here 2
were 1
Is it possible to merge df with the newly aggregated table gb such that I create a new column in df, containing the corresponding values from gb? Like this:
a b c nc
0 1 4 wish 1
1 2 5 you 1
2 3 6 were 1
3 13 6 here 2
4 15 6 here 2
I tried doing the simplest thing:
df.merge(gb, on=['b','c'])
But this gives the error:
KeyError: 'b'
Which makes sense because the grouped table has a Multi-index and b is not a column. So my question is two-fold:
Can I transform the multi-index of the gb DataFrame back into columns (so that it has the b and c column)?
Can I merge df with gb on the column names?

Whenever you want to add some aggregated column from groupby operation back to the df you should be using transform, this produces a Series with its index aligned with your orig df:
In [4]:
df['nc'] = df.groupby(['b','c'])['a'].transform(pd.Series.nunique)
df
Out[4]:
a b c nc
0 1 4 wish 1
1 2 5 you 1
2 3 6 were 1
3 13 6 here 2
4 15 6 here 2
There is no need to reset the index or perform an additional merge.

There's a simple way of doing this using reset_index().
df.merge(gb.reset_index(), on=['b','c'])
gives you
a_x b c a_y
0 1 4 wish 1
1 2 5 you 1
2 3 6 were 1
3 13 6 here 2
4 15 6 here 2