Pandas distinct count column - python-3.x

Inspired by this post, I would like to get a distinct count of a value in a data frame per a grouping and create a column with the distinct count values in the data frame.
Like this:
Original data frame:
import pandas as pd
df = pd.DataFrame(
{'A' : ['foo', 'foo', 'foo', 'foo',
'bar', 'bar', 'bar', 'bar'],
'B' : ['foo', 'fo', 'foo', 'foo',
'bar', 'bar', 'ba', 'ba'],
'C' : [2, 4, 4, 2, 5, 4, 3, 2]})
df
A B C
0 foo foo 2
1 foo fo 4
2 foo foo 4
3 foo foo 2
4 bar bar 5
5 bar bar 4
6 bar ba 3
7 bar ba 2
Method from linked post applied:
df=df.groupby(['A','B'])['C'].apply(lambda x: len(x.unique()))
df
Result per linked post method:
A B
bar ba 2
bar 2
foo fo 1
foo 2
Name: C, dtype: int64
Desired result:
A B C Distinct Count of C per A and B
0 foo foo 2 2
1 foo fo 4 1
2 foo foo 4 2
3 foo foo 2 2
4 bar bar 5 2
5 bar bar 4 2
6 bar ba 3 2
7 bar ba 2 2
Looking at the first row, the combination of "foo" in "A" and "foo" in "B" has 2 unique values associated with it (2 and 4), resulting in a 2 in each row for that combination of values for columns A and B.
Thank in advance!


Use transform instead of apply because it return column with the same size as original, I couldn't find documentation on the original pandas site for that, but from help:
transform(func, *args, **kwargs) method of
pandas.core.groupby.SeriesGroupBy instance
Call function producing a like-indexed Series on each group and return
a Series with the transformed values
df['Distinct Count of C per A and B'] = df.groupby(['A','B'])['C'].transform(lambda x: len(x.unique()))
In [1495]: df
Out[1495]:
A B C Distinct Count of C per A and B
0 foo foo 2 2
1 foo fo 4 1
2 foo foo 4 2
3 foo foo 2 2
4 bar bar 5 2
5 bar bar 4 2
6 bar ba 3 2
7 bar ba 2 2

Related

Split pandas Dataframe into 4 parts with stratified sampling

I want to split a Dataframe into 4 parts with stratified sampling.
Make sure all categories form column 'B' Should present in each chunk. If any category is not having sufficient records for all chunks, copy same record into remaining chunks.
df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
'foo', 'bar', 'foo', 'foo',
'foo', 'bar', 'foo', 'bar',
'foo', 'bar', 'foo', 'foo', 'bar'],
'B' : ['one', 'one', 'two', 'three',
'two', 'two', 'one', 'three',
'one', 'one', 'two', 'three',
'two', 'two', 'one', 'three', 'four'],
'C' : np.random.randn(17), 'D' : np.random.randn(17)})
print(df)
A B C D
0 foo one 0.960627 0.318723
1 bar one 0.269439 -0.945565
2 foo two 0.210376 0.765680
3 bar three -0.375095 -1.617334
4 foo two -1.910716 -0.532117
5 bar two -0.277426 0.019717
6 foo one -0.260074 1.384464
7 foo three 0.072119 -1.077725
8 foo one 0.093446 -0.683513
9 bar one -0.154885 -1.453996
10 foo two -1.258207 1.406615
11 bar three -0.003332 -0.083092
12 foo two 1.250562 0.519337
13 bar two -0.837681 -1.465363
14 foo one -0.403992 -0.133496
15 foo three -0.757623 -0.459532
16 bar four -2.071840 0.802953
Output should be like below
(All categories from 'B' column should present in each chunk. Index doesn't matter)
A B C D
0 foo one 0.200466 -0.394136
2 foo two 0.086008 -0.528286
3 bar three -1.979613 -1.345405
8 foo one -1.195563 -0.832880
15 foo three -0.737060 -0.437047
16 bar four -2.071840 0.802953
A B C D
1 bar one 1.177119 0.693766
4 foo two 0.452803 -0.595433
7 foo three 1.285687 1.107021
12 foo two 1.746976 1.449390
16 bar four -2.071840 0.802953
A B C D
6 foo one -0.095485 0.129541
5 bar two 0.803417 -0.219461
7 foo three 1.285687 1.107021
13 bar two 1.166246 -1.711505
16 bar four -2.071840 0.802953
A B C D
9 bar one 2.001238 -0.283411
10 foo two 0.865580 0.052533
11 bar three -0.437604 -0.652073
14 foo one -0.655985 -0.942792
16 bar four -2.071840 0.802953

This may help:
df1, df2, df3, df4 = np.array_split(x_train, 4)
from: Split large Dataframe into smaller equal dataframes

Select row by max of a column Pandas Python [duplicate]

How can I perform aggregation with Pandas?
No DataFrame after aggregation! What happened?
How can I aggregate mainly strings columns (to lists, tuples, strings with separator)?
How can I aggregate counts?
How can I create a new column filled by aggregated values?
I've seen these recurring questions asking about various faces of the pandas aggregate functionality.
Most of the information regarding aggregation and its various use cases today is fragmented across dozens of badly worded, unsearchable posts.
The aim here is to collate some of the more important points for posterity.
This Q&A is meant to be the next instalment in a series of helpful user-guides:
How to pivot a dataframe,
Pandas concat
How do I operate on a DataFrame with a Series for every column?
Pandas Merging 101
Please note that this post is not meant to be a replacement for the documentation about aggregation and about groupby, so please read that as well!

Question 1
How can I perform aggregation with Pandas?
Expanded aggregation documentation.
Aggregating functions are the ones that reduce the dimension of the returned objects. It means output Series/DataFrame have less or same rows like original.
Some common aggregating functions are tabulated below:
Function Description
mean() Compute mean of groups
sum() Compute sum of group values
size() Compute group sizes
count() Compute count of group
std() Standard deviation of groups
var() Compute variance of groups
sem() Standard error of the mean of groups
describe() Generates descriptive statistics
first() Compute first of group values
last() Compute last of group values
nth() Take nth value, or a subset if n is a list
min() Compute min of group values
max() Compute max of group values
np.random.seed(123)
df = pd.DataFrame({'A' : ['foo', 'foo', 'bar', 'foo', 'bar', 'foo'],
'B' : ['one', 'two', 'three','two', 'two', 'one'],
'C' : np.random.randint(5, size=6),
'D' : np.random.randint(5, size=6),
'E' : np.random.randint(5, size=6)})
print (df)
A B C D E
0 foo one 2 3 0
1 foo two 4 1 0
2 bar three 2 1 1
3 foo two 1 0 3
4 bar two 3 1 4
5 foo one 2 1 0
Aggregation by filtered columns and Cython implemented functions:
df1 = df.groupby(['A', 'B'], as_index=False)['C'].sum()
print (df1)
A B C
0 bar three 2
1 bar two 3
2 foo one 4
3 foo two 5
An aggregate function is used for all columns without being specified in the groupby function, here the A, B columns:
df2 = df.groupby(['A', 'B'], as_index=False).sum()
print (df2)
A B C D E
0 bar three 2 1 1
1 bar two 3 1 4
2 foo one 4 4 0
3 foo two 5 1 3
You can also specify only some columns used for aggregation in a list after the groupby function:
df3 = df.groupby(['A', 'B'], as_index=False)['C','D'].sum()
print (df3)
A B C D
0 bar three 2 1
1 bar two 3 1
2 foo one 4 4
3 foo two 5 1
Same results by using function DataFrameGroupBy.agg:
df1 = df.groupby(['A', 'B'], as_index=False)['C'].agg('sum')
print (df1)
A B C
0 bar three 2
1 bar two 3
2 foo one 4
3 foo two 5
df2 = df.groupby(['A', 'B'], as_index=False).agg('sum')
print (df2)
A B C D E
0 bar three 2 1 1
1 bar two 3 1 4
2 foo one 4 4 0
3 foo two 5 1 3
For multiple functions applied for one column use a list of tuples - names of new columns and aggregated functions:
df4 = (df.groupby(['A', 'B'])['C']
.agg([('average','mean'),('total','sum')])
.reset_index())
print (df4)
A B average total
0 bar three 2.0 2
1 bar two 3.0 3
2 foo one 2.0 4
3 foo two 2.5 5
If want to pass multiple functions is possible pass list of tuples:
df5 = (df.groupby(['A', 'B'])
.agg([('average','mean'),('total','sum')]))
print (df5)
C D E
average total average total average total
A B
bar three 2.0 2 1.0 1 1.0 1
two 3.0 3 1.0 1 4.0 4
foo one 2.0 4 2.0 4 0.0 0
two 2.5 5 0.5 1 1.5 3
Then get MultiIndex in columns:
print (df5.columns)
MultiIndex(levels=[['C', 'D', 'E'], ['average', 'total']],
labels=[[0, 0, 1, 1, 2, 2], [0, 1, 0, 1, 0, 1]])
And for converting to columns, flattening MultiIndex use map with join:
df5.columns = df5.columns.map('_'.join)
df5 = df5.reset_index()
print (df5)
A B C_average C_total D_average D_total E_average E_total
0 bar three 2.0 2 1.0 1 1.0 1
1 bar two 3.0 3 1.0 1 4.0 4
2 foo one 2.0 4 2.0 4 0.0 0
3 foo two 2.5 5 0.5 1 1.5 3
Another solution is pass list of aggregate functions, then flatten MultiIndex and for another columns names use str.replace:
df5 = df.groupby(['A', 'B']).agg(['mean','sum'])
df5.columns = (df5.columns.map('_'.join)
.str.replace('sum','total')
.str.replace('mean','average'))
df5 = df5.reset_index()
print (df5)
A B C_average C_total D_average D_total E_average E_total
0 bar three 2.0 2 1.0 1 1.0 1
1 bar two 3.0 3 1.0 1 4.0 4
2 foo one 2.0 4 2.0 4 0.0 0
3 foo two 2.5 5 0.5 1 1.5 3
If want specified each column with aggregated function separately pass dictionary:
df6 = (df.groupby(['A', 'B'], as_index=False)
.agg({'C':'sum','D':'mean'})
.rename(columns={'C':'C_total', 'D':'D_average'}))
print (df6)
A B C_total D_average
0 bar three 2 1.0
1 bar two 3 1.0
2 foo one 4 2.0
3 foo two 5 0.5
You can pass custom function too:
def func(x):
return x.iat[0] + x.iat[-1]
df7 = (df.groupby(['A', 'B'], as_index=False)
.agg({'C':'sum','D': func})
.rename(columns={'C':'C_total', 'D':'D_sum_first_and_last'}))
print (df7)
A B C_total D_sum_first_and_last
0 bar three 2 2
1 bar two 3 2
2 foo one 4 4
3 foo two 5 1
Question 2
No DataFrame after aggregation! What happened?
Aggregation by two or more columns:
df1 = df.groupby(['A', 'B'])['C'].sum()
print (df1)
A B
bar three 2
two 3
foo one 4
two 5
Name: C, dtype: int32
First check the Index and type of a Pandas object:
print (df1.index)
MultiIndex(levels=[['bar', 'foo'], ['one', 'three', 'two']],
labels=[[0, 0, 1, 1], [1, 2, 0, 2]],
names=['A', 'B'])
print (type(df1))
<class 'pandas.core.series.Series'>
There are two solutions for how to get MultiIndex Series to columns:
add parameter as_index=False
df1 = df.groupby(['A', 'B'], as_index=False)['C'].sum()
print (df1)
A B C
0 bar three 2
1 bar two 3
2 foo one 4
3 foo two 5
use Series.reset_index:
df1 = df.groupby(['A', 'B'])['C'].sum().reset_index()
print (df1)
A B C
0 bar three 2
1 bar two 3
2 foo one 4
3 foo two 5
If group by one column:
df2 = df.groupby('A')['C'].sum()
print (df2)
A
bar 5
foo 9
Name: C, dtype: int32
... get Series with Index:
print (df2.index)
Index(['bar', 'foo'], dtype='object', name='A')
print (type(df2))
<class 'pandas.core.series.Series'>
And the solution is the same like in the MultiIndex Series:
df2 = df.groupby('A', as_index=False)['C'].sum()
print (df2)
A C
0 bar 5
1 foo 9
df2 = df.groupby('A')['C'].sum().reset_index()
print (df2)
A C
0 bar 5
1 foo 9
Question 3
How can I aggregate mainly strings columns (to lists, tuples, strings with separator)?
df = pd.DataFrame({'A' : ['a', 'c', 'b', 'b', 'a', 'c', 'b'],
'B' : ['one', 'two', 'three','two', 'two', 'one', 'three'],
'C' : ['three', 'one', 'two', 'two', 'three','two', 'one'],
'D' : [1,2,3,2,3,1,2]})
print (df)
A B C D
0 a one three 1
1 c two one 2
2 b three two 3
3 b two two 2
4 a two three 3
5 c one two 1
6 b three one 2
Instead of an aggregation function, it is possible to pass list, tuple, set for converting the column:
df1 = df.groupby('A')['B'].agg(list).reset_index()
print (df1)
A B
0 a [one, two]
1 b [three, two, three]
2 c [two, one]
An alternative is use GroupBy.apply:
df1 = df.groupby('A')['B'].apply(list).reset_index()
print (df1)
A B
0 a [one, two]
1 b [three, two, three]
2 c [two, one]
For converting to strings with a separator, use .join only if it is a string column:
df2 = df.groupby('A')['B'].agg(','.join).reset_index()
print (df2)
A B
0 a one,two
1 b three,two,three
2 c two,one
If it is a numeric column, use a lambda function with astype for converting to strings:
df3 = (df.groupby('A')['D']
.agg(lambda x: ','.join(x.astype(str)))
.reset_index())
print (df3)
A D
0 a 1,3
1 b 3,2,2
2 c 2,1
Another solution is converting to strings before groupby:
df3 = (df.assign(D = df['D'].astype(str))
.groupby('A')['D']
.agg(','.join).reset_index())
print (df3)
A D
0 a 1,3
1 b 3,2,2
2 c 2,1
For converting all columns, don't pass a list of column(s) after groupby.
There isn't any column D, because automatic exclusion of 'nuisance' columns. It means all numeric columns are excluded.
df4 = df.groupby('A').agg(','.join).reset_index()
print (df4)
A B C
0 a one,two three,three
1 b three,two,three two,two,one
2 c two,one one,two
So it's necessary to convert all columns into strings, and then get all columns:
df5 = (df.groupby('A')
.agg(lambda x: ','.join(x.astype(str)))
.reset_index())
print (df5)
A B C D
0 a one,two three,three 1,3
1 b three,two,three two,two,one 3,2,2
2 c two,one one,two 2,1
Question 4
How can I aggregate counts?
df = pd.DataFrame({'A' : ['a', 'c', 'b', 'b', 'a', 'c', 'b'],
'B' : ['one', 'two', 'three','two', 'two', 'one', 'three'],
'C' : ['three', np.nan, np.nan, 'two', 'three','two', 'one'],
'D' : [np.nan,2,3,2,3,np.nan,2]})
print (df)
A B C D
0 a one three NaN
1 c two NaN 2.0
2 b three NaN 3.0
3 b two two 2.0
4 a two three 3.0
5 c one two NaN
6 b three one 2.0
Function GroupBy.size for size of each group:
df1 = df.groupby('A').size().reset_index(name='COUNT')
print (df1)
A COUNT
0 a 2
1 b 3
2 c 2
Function GroupBy.count excludes missing values:
df2 = df.groupby('A')['C'].count().reset_index(name='COUNT')
print (df2)
A COUNT
0 a 2
1 b 2
2 c 1
This function should be used for multiple columns for counting non-missing values:
df3 = df.groupby('A').count().add_suffix('_COUNT').reset_index()
print (df3)
A B_COUNT C_COUNT D_COUNT
0 a 2 2 1
1 b 3 2 3
2 c 2 1 1
A related function is Series.value_counts. It returns the size of the object containing counts of unique values in descending order, so that the first element is the most frequently-occurring element. It excludes NaNs values by default.
df4 = (df['A'].value_counts()
.rename_axis('A')
.reset_index(name='COUNT'))
print (df4)
A COUNT
0 b 3
1 a 2
2 c 2
If you want same output like using function groupby + size, add Series.sort_index:
df5 = (df['A'].value_counts()
.sort_index()
.rename_axis('A')
.reset_index(name='COUNT'))
print (df5)
A COUNT
0 a 2
1 b 3
2 c 2
Question 5
How can I create a new column filled by aggregated values?
Method GroupBy.transform returns an object that is indexed the same (same size) as the one being grouped.
See the Pandas documentation for more information.
np.random.seed(123)
df = pd.DataFrame({'A' : ['foo', 'foo', 'bar', 'foo', 'bar', 'foo'],
'B' : ['one', 'two', 'three','two', 'two', 'one'],
'C' : np.random.randint(5, size=6),
'D' : np.random.randint(5, size=6)})
print (df)
A B C D
0 foo one 2 3
1 foo two 4 1
2 bar three 2 1
3 foo two 1 0
4 bar two 3 1
5 foo one 2 1
df['C1'] = df.groupby('A')['C'].transform('sum')
df['C2'] = df.groupby(['A','B'])['C'].transform('sum')
df[['C3','D3']] = df.groupby('A')['C','D'].transform('sum')
df[['C4','D4']] = df.groupby(['A','B'])['C','D'].transform('sum')
print (df)
A B C D C1 C2 C3 D3 C4 D4
0 foo one 2 3 9 4 9 5 4 4
1 foo two 4 1 9 5 9 5 5 1
2 bar three 2 1 5 2 5 2 2 1
3 foo two 1 0 9 5 9 5 5 1
4 bar two 3 1 5 3 5 2 3 1
5 foo one 2 1 9 4 9 5 4 4

If you are coming from an R or SQL background, here are three examples that will teach you everything you need to do aggregation the way you are already familiar with:
Let us first create a Pandas dataframe
import pandas as pd
df = pd.DataFrame({'key1' : ['a','a','a','b','a'],
'key2' : ['c','c','d','d','e'],
'value1' : [1,2,2,3,3],
'value2' : [9,8,7,6,5]})
df.head(5)
Here is how the table we created looks like:
key1
key2
value1
value2
a
c
1
9
a
c
2
8
a
d
2
7
b
d
3
6
a
e
3
5
1. Aggregating With Row Reduction Similar to SQL Group By
1.1 If Pandas version >=0.25
Check your Pandas version by running print(pd.__version__). If your Pandas version is 0.25 or above then the following code will work:
df_agg = df.groupby(['key1','key2']).agg(mean_of_value_1=('value1', 'mean'),
sum_of_value_2=('value2', 'sum'),
count_of_value1=('value1','size')
).reset_index()
df_agg.head(5)
The resulting data table will look like this:
key1
key2
mean_of_value1
sum_of_value2
count_of_value1
a
c
1.5
17
2
a
d
2.0
7
1
a
e
3.0
5
1
b
d
3.0
6
1
The SQL equivalent of this is:
SELECT
key1
,key2
,AVG(value1) AS mean_of_value_1
,SUM(value2) AS sum_of_value_2
,COUNT(*) AS count_of_value1
FROM
df
GROUP BY
key1
,key2
1.2 If Pandas version <0.25
If your Pandas version is older than 0.25 then running the above code will give you the following error:
TypeError: aggregate() missing 1 required positional argument: 'arg'
Now to do the aggregation for both value1 and value2, you will run this code:
df_agg = df.groupby(['key1','key2'],as_index=False).agg({'value1':['mean','count'],'value2':'sum'})
df_agg.columns = ['_'.join(col).strip() for col in df_agg.columns.values]
df_agg.head(5)
The resulting table will look like this:
key1
key2
value1_mean
value1_count
value2_sum
a
c
1.5
2
17
a
d
2.0
1
7
a
e
3.0
1
5
b
d
3.0
1
6
Renaming the columns needs to be done separately using the below code:
df_agg.rename(columns={"value1_mean" : "mean_of_value1",
"value1_count" : "count_of_value1",
"value2_sum" : "sum_of_value2"
}, inplace=True)
2. Create a Column Without Reduction in Rows (EXCEL - SUMIF, COUNTIF)
If you want to do a SUMIF, COUNTIF, etc., like how you would do in Excel where there is no reduction in rows, then you need to do this instead.
df['Total_of_value1_by_key1'] = df.groupby('key1')['value1'].transform('sum')
df.head(5)
The resulting data frame will look like this with the same number of rows as the original:
key1
key2
value1
value2
Total_of_value1_by_key1
a
c
1
9
8
a
c
2
8
8
a
d
2
7
8
b
d
3
6
3
a
e
3
5
8
3. Creating a RANK Column ROW_NUMBER() OVER (PARTITION BY ORDER BY)
Finally, there might be cases where you want to create a rank column which is the SQL equivalent of ROW_NUMBER() OVER (PARTITION BY key1 ORDER BY value1 DESC, value2 ASC).
Here is how you do that.
df['RN'] = df.sort_values(['value1','value2'], ascending=[False,True]) \
.groupby(['key1']) \
.cumcount() + 1
df.head(5)
Note: we make the code multi-line by adding \ at the end of each line.
Here is how the resulting data frame looks like:
key1
key2
value1
value2
RN
a
c
1
9
4
a
c
2
8
3
a
d
2
7
2
b
d
3
6
1
a
e
3
5
1
In all the examples above, the final data table will have a table structure and won't have the pivot structure that you might get in other syntaxes.
Other aggregating operators:
mean() Compute mean of groups
sum() Compute sum of group values
size() Compute group sizes
count() Compute count of group
std() Standard deviation of groups
var() Compute variance of groups
sem() Standard error of the mean of groups
describe() Generates descriptive statistics
first() Compute first of group values
last() Compute last of group values
nth() Take nth value, or a subset if n is a list
min() Compute min of group values
max() Compute max of group values

Getting all rows where for column 'C' the entry is larger than the preceding element in column 'C'

How can I select all rows of a data frame where a condition is met according to a column, which has to do with the relationship between every 2 entries of that column. To give the specific example, lets say I have a DataFrame:
>>>df = pd.DataFrame({'A': [ 1, 2, 3, 4],
'B':['spam', 'ham', 'egg', 'foo'],
'C':[4, 5, 3, 4]})
>>> df
A B C
0 1 spam 4
1 2 ham 5
2 3 egg 3
3 4 foo 4
>>>df2 = df[ return every row of df where C[i] > C[i-1] ]
>>> df2
A B C
1 2 ham 5
3 4 foo 4
There is plenty of great information about slicing and indexing in the pandas docs and here, but this is a bit more complicated, I think. I could also be going about it wrong. What I'm looking for is the rows of data where the value stored in C is no longer monotonously declining.
Any help is appreciated!

Use boolean indexing with compare by shifted column values:
print (df[df['C'] > df['C'].shift()])
A B C
1 2 ham 5
3 4 foo 4
Detail:
print (df['C'] > df['C'].shift())
0 False
1 True
2 False
3 True
Name: C, dtype: bool
If want all monotonously declining rows compare diff of column:
print (df[df['C'].diff() > 0])
A B C
1 2 ham 5
3 4 foo 4

Pandas: Calculating value of difference between current column value and next column value depending if it meets criteria at a different column

I have a dataframe:
df = pd.DataFrame.from_items([('A', [10, 'foo']), ('B', [440, 'foo']), ('C', [790, 'bar']), ('D', [800, 'bar']), ('E', [7000, 'foo'])], orient='index', columns=['position', 'foobar'])
Which looks like the below:
position foobar
A 10 foo
B 440 foo
C 790 bar
D 800 bar
E 7000 foo
I would like to know the difference between each position and the next position that has the opposite value in the foobar column. Normally I would use the shift method to move down the position column:
df[comparisonCol].shift(-1) - df[comparisonCol]
but as I am using the foobar column to decide which position is applicable, I am not sure how to do this.
The result should look like:
position foobar difference
A 10 foo 780
B 440 foo 350
C 790 bar 6210
D 800 bar 6200
E 7000 foo NaN

I think you need if unique values in foobar are only 2, so is possible shift between groups in a Series:
#identify consecutive groups
a = df['foobar'].ne(df['foobar'].shift()).cumsum()
print (a)
A 1
B 1
C 2
D 2
E 3
Name: foobar, dtype: int32
#get first value by a of position column
b = df.groupby(a)['position'].first()
print (b)
foobar
1 10
2 790
3 7000
Name: position, dtype: int64
#subtract mapped value, but for next group is added 1 to a Series
df['difference'] = a.add(1).map(b) - df['position']
print (df)
position foobar difference
A 10 foo 780.0
B 440 foo 350.0
C 790 bar 6210.0
D 800 bar 6200.0
E 7000 foo NaN
Detail:
print (a.add(1).map(b))
A 790.0
B 790.0
C 7000.0
D 7000.0
E NaN
Name: foobar, dtype: float64

Pandas Conditional Groupby Count Part 2

Given this problem:
Pandas conditional groupby count
I would like the result to be this instead:
A D Dcount
0 foo 2 2
1 foo 4 2
2 foo 4 2
3 foo 2 2
4 bar 5 NaN
5 bar 4 NaN
6 bar 3 NaN
7 bar 2 NaN
What I mean is, if 2 conditions are met (column A = 'foo' and column B = 2),
I'd like for there to be the distinct count of such rows (2) in the Dcount column for all rows of column A = 'foo'.
Can this be modified to allow for the desired result?
import pandas as pd
df = pd.DataFrame(
{'A' : ['foo', 'foo', 'foo', 'foo',
'bar', 'bar', 'bar', 'bar'],
'D' : [2, 4, 4, 2, 5, 4, 3, 2]})
#First, I filter
df2=df.loc[(df['A']=='foo')&(df['D']==2)]
#Then, I use groupby and lambda x to count
df['Dcount']=df2.groupby(['D'])['D'].transform(lambda x: x.count())
df
Thanks in advance!

You can use where from numpy in a one-liner:
import numpy as np
df['Dcount'] = np.where(df['A']=='foo', sum((df.A=='foo') & (df.D==2)), np.NaN)
#In [34]: df
#Out[34]:
# A D Dcount
#0 foo 2 2
#1 foo 4 2
#2 foo 4 2
#3 foo 2 2
#4 bar 5 NaN
#5 bar 4 NaN
#6 bar 3 NaN
#7 bar 2 NaN

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