I have this pandas dataframe:
SourceDomain 1 2 3
0 www.theguardian.com profile.theguardian.com 1 Directed
1 www.theguardian.com membership.theguardian.com 2 Directed
2 www.theguardian.com subscribe.theguardian.com 3 Directed
3 www.theguardian.com www.google.co.uk 4 Directed
4 www.theguardian.com jobs.theguardian.com 5 Directed
I would like to add a new column which is a pandas series created like this:
Weights = Weights.value_counts()
However, when I try to add the new column using edgesFile[4] = Weights it fills it with NA instead of the values:
SourceDomain 1 2 3 4
0 www.theguardian.com profile.theguardian.com 1 Directed NaN
1 www.theguardian.com membership.theguardian.com 2 Directed NaN
2 www.theguardian.com subscribe.theguardian.com 3 Directed NaN
3 www.theguardian.com www.google.co.uk 4 Directed NaN
4 www.theguardian.com jobs.theguardian.com 5 Directed NaN
How can I add the new column keeping the values?
Thanks?
Dani
You are getting NaNs because the index of Weights does not match up with the index of edgesFile. If you want Pandas to ignore Weights.index and just paste the values in order then pass the underlying NumPy array instead:
edgesFile[4] = Weights.values
Here is an example which demonstrates the difference:
In [14]: df = pd.DataFrame(np.arange(4)*10, index=list('ABCD'))
In [15]: df
Out[15]:
0
A 0
B 10
C 20
D 30
In [16]: s = pd.Series(np.arange(4), index=list('CDEF'))
In [17]: s
Out[17]:
C 0
D 1
E 2
F 3
dtype: int64
Here we see Pandas aligning the index:
In [18]: df[4] = s
In [19]: df
Out[19]:
0 4
A 0 NaN
B 10 NaN
C 20 0
D 30 1
Here, Pandas simply pastes the values in s into the column:
In [20]: df[4] = s.values
In [21]: df
Out[21]:
0 4
A 0 0
B 10 1
C 20 2
D 30 3
This is small example of your question:
You can add new column with a column name in existing DataFrame
>>> df = DataFrame([[1,2,3],[4,5,6]], columns = ['A', 'B', 'C'])
>>> df
A B C
0 1 2 3
1 4 5 6
>>> s = Series([7,8])
>>> s
0 7
1 8
2 9
>>> df['D']=s
>>> df
A B C D
0 1 2 3 7
1 4 5 6 8
Or, You can make DataFrame from Series and concat then
>>> df = DataFrame([[1,2,3],[4,5,6]])
>>> df
0 1 2
0 1 2 3
1 4 5 6
>>> s = DataFrame(Series([7,8]), columns=['4']) # if you don't provide column name, default name will be 0
>>> s
0
0 7
1 8
>>> df = pd.concat([df,s], axis=1)
>>> df
0 1 2 0
0 1 2 3 7
1 4 5 6 8
Hope this will help
Related
this is my data and i want to find the min value of selected columns(a,b,c,d) in each row then calculate the difference between that and dd. I need to ignore 0 in rows, I mean in the first row i need to find 8
need to ignore 0 in rows
Then just replace it with nan, consider following simple example
import numpy as np
import pandas as pd
df = pd.DataFrame({"A":[1,2,0],"B":[3,5,7],"C":[7,0,7]})
df.replace(0,np.nan).apply(min)
df["minvalue"] = df.replace(0,np.nan).apply("min",axis=1)
print(df)
gives output
A B C minvalue
0 1 3 7 1.0
1 2 5 0 2.0
2 0 7 7 7.0
You can use pandas.apply with axis=1 and all column ['a','b','c','d'] convert to Series then replace 0 with +inf and find min. At the end compute diff min with colmun 'dd'.
import numpy as np
df['min_dd'] = df.apply(lambda row: min(pd.Series(row[['a','b','c','d']]).replace(0,np.inf)) - row['d'], axis=1)
print(df)
a b c d dd min_dd
0 0 15 0 8 6 2.0 # min_without_zero : 8 , dd : 6 -> 8-6=2
1 2 0 5 3 2 0.0 # min_without_zero : 2 , dd : 2 -> 2-2=0
2 5 3 3 0 2 1.0 # 3 - 2
3 0 2 3 4 2 0.0 # 2 - 2
You can try
cols = ['a','b','c','d']
df['res'] = df[cols][df[cols].ne(0)].min(axis=1) - df['dd']
print(df)
a b c d dd res
0 0 15 0 8 6 2.0
1 2 0 5 3 2 0.0
2 5 3 3 0 2 1.0
3 2 3 4 4 2 0.0
I was doing some coding and realized something, I think there is an easier way of doing this.
So I have a DataFrame like this:
>>> df = pd.DataFrame({'a': [1, 'A', 2, 'A'], 'b': ['A', 3, 'A', 4]})
a b
0 1 A
1 A 3
2 2 A
3 A 4
And I want to remove all of the As from the data, but I also want to squeeze in the DataFrame, what I mean by squeezing in the DataFrame is to have a result of this:
a b
0 1 3
1 2 4
I have a solution as follows:
a = df['a'][df['a'] != 'A']
b = df['b'][df['b'] != 'A']
df2 = pd.DataFrame({'a': a.tolist(), 'b': b.tolist()})
print(df2)
Which works, but I seem to think there is an easier way, I've stopped coding for a while so not so bright anymore...
Note:
All columns have the same amount of As, there is no problem there.
You can try boolean indexing with loc to remove the A values:
pd.DataFrame({c: df.loc[df[c] != 'A', c].tolist() for c in df})
Result:
a b
0 1 3
1 2 4
This would do:
In [1513]: df.replace('A', np.nan).apply(lambda x: pd.Series(x.dropna().to_numpy()))
Out[1513]:
a b
0 1.0 3.0
1 2.0 4.0
We use can df.melt then filter out 'A' values then df.pivot
out = df.melt().query("value!='A'")
out.index = out.groupby('variable')['variable'].cumcount()
out.pivot(columns='variable', values='value').rename_axis(columns=None)
a b
0 1 3
1 2 4
Details
out = df.melt().query("value!='A'")
variable value
0 a 1
2 a 2
5 b 3
7 b 4
# We set this as index so it helps in `df.pivot`
out.groupby('variable')['variable'].cumcount()
0 0
2 1
5 0
7 1
dtype: int64
out.pivot(columns='variable', values='value').rename_axis(columns=None)
a b
0 1 3
1 2 4
Another alternative
df = df.mask(df.eq('A'))
out = df.stack()
pd.DataFrame(out.groupby(level=1).agg(list).to_dict())
a b
0 1 3
1 2 4
Details
df = df.mask(df.eq('A'))
a b
0 1 NaN
1 NaN 3
2 2 NaN
3 NaN 4
out = df.stack()
0 a 1
1 b 3
2 a 2
3 b 4
dtype: object
pd.DataFrame(out.groupby(level=1).agg(list).to_dict())
a b
0 1 3
1 2 4
New to Pandas, not very sure how the 3D DataFrame works. My dataframe, called 'new' looks like this:
unique cat numerical
a b c d e f
0 0 1 2 3 4 5
1 0 1 2 3 4 5
I want to insert column 'z' so that it ends up like this:
unique cat numerical
a b z c d e f
0 0 1 9 2 3 4 5
1 0 1 9 2 3 4 5
I successfully made a new column after slicing out 'unique' from my dataframe:
Doing this:
new_column = new.loc[:,'unique'].assign(z=pd.Series([9,9]).values)
Gets me this:
a b z
0 0 1 9
1 0 1 9
However I have no idea how to put it back into the dataframe. I tried:
new['unique'] = new_column
But I've since found out that it just tries to replace all the values in all the rows and columns found under 'unique', like this:
new['unique'] = 'a'
Gets:
unique cat numerical
a b c d e f
0 a a 2 3 4 5
1 a a 2 3 4 5
And using .loc gets this instead:
unique cat numerical
a b c d e f
0 NaN NaN 2 3 4 5
1 NaN NaN 2 3 4 5
Here's my full code:
import pandas as pd
import numpy as np
data=[[0,1,2,3,4,5],[0,1,2,3,4,5]]
datatypes=np.array(['unique','unique','cat','cat','numerical','numerical'])
columnnames=np.array(['a','b','c','d','e','f'])
new = pd.DataFrame(data=data, columns=pd.MultiIndex.from_tuples(zip(datatypes,columnnames)))
print('new: ')
print(new)
new_column = new.loc[:,'unique'].assign(z=pd.Series([9,9]).values)
print('\nnew column:')
print(new_column)
new.loc[:,'unique'] = new_column
print('\nattempt 1:')
print(new)
new['unique'] = new_column
print('\nattempt 2:')
print(new)
One way to do this:
# Create your new multiindexed column:
new['unique','z'] = 9
# Re-order your columns in your desired order:
new = new[['unique', 'cat', 'numerical']]
>>> new
unique cat numerical
a b z c d e f
0 0 1 9 2 3 4 5
1 0 1 9 2 3 4 5
I have a DataFrame with column names in the shape of x.y, where I would like to sum up all columns with the same value on x without having to explicitly name them. That is, the value of column_name.split(".")[0] should determine their group. Here's an example:
import pandas as pd
df = pd.DataFrame({'x.1': [1,2,3,4], 'x.2': [5,4,3,2], 'y.8': [19,2,1,3], 'y.92': [10,9,2,4]})
df
Out[3]:
x.1 x.2 y.8 y.92
0 1 5 19 10
1 2 4 2 9
2 3 3 1 2
3 4 2 3 4
The result should be the same as this operation, only I shouldn't have to explicitly list the column names and how they should group.
pd.DataFrame({'x': df[['x.1', 'x.2']].sum(axis=1), 'y': df[['y.8', 'y.92']].sum(axis=1)})
x y
0 6 29
1 6 11
2 6 3
3 6 7
Another option, you can extract the prefix from the column names and use it as a group variable:
df.groupby(by = df.columns.str.split('.').str[0], axis = 1).sum()
# x y
#0 6 29
#1 6 11
#2 6 3
#3 6 7
You can first create Multiindex by split and then groupby by first level and aggregate sum:
df.columns = df.columns.str.split('.', expand=True)
print (df)
x y
1 2 8 92
0 1 5 19 10
1 2 4 2 9
2 3 3 1 2
3 4 2 3 4
df = df.groupby(axis=1, level=0).sum()
print (df)
x y
0 6 29
1 6 11
2 6 3
3 6 7
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