Use groupby and merge to create new column in pandas - python

So I have a pandas dataframe that looks something like this.
name is_something
0 a 0
1 b 1
2 c 0
3 c 1
4 a 1
5 b 0
6 a 1
7 c 0
8 a 1
Is there a way to use groupby and merge to create a new column that gives the number of times a name appears with an is_something value of 1 in the whole dataframe? The updated dataframe would look like this:
name is_something no_of_times_is_something_is_1
0 a 0 3
1 b 1 1
2 c 0 1
3 c 1 1
4 a 1 3
5 b 0 1
6 a 1 3
7 c 0 1
8 a 1 3
I know you can just loop through the dataframe to do this but I'm looking for a more efficient way because the dataset I'm working with is quite large. Thanks in advance!


If there are only 0 and 1 values in is_something column only use sum with GroupBy.transform for new column filled by aggregate values:
df['new'] = df.groupby('name')['is_something'].transform('sum')
print (df)
name is_something new
0 a 0 3
1 b 1 1
2 c 0 1
3 c 1 1
4 a 1 3
5 b 0 1
6 a 1 3
7 c 0 1
8 a 1 3
If possible multiple values first compare by 1, convert to integer and then use transform with sum:
df['new'] = df['is_something'].eq(1).view('i1').groupby(df['name']).transform('sum')


Or we just map it
df['New']=df.name.map(df.query('is_something ==1').groupby('name')['is_something'].sum())
df
name is_something New
0 a 0 3
1 b 1 1
2 c 0 1
3 c 1 1
4 a 1 3
5 b 0 1
6 a 1 3
7 c 0 1
8 a 1 3


You could do:
df['new'] = df.groupby('name')['is_something'].transform(lambda xs: xs.eq(1).sum())
print(df)
Output
name is_something new
0 a 0 3
1 b 1 1
2 c 0 1
3 c 1 1
4 a 1 3
5 b 0 1
6 a 1 3
7 c 0 1
8 a 1 3

Related

Set value when row is maximum in group by - Python Pandas

I am trying to create a column (is_max) that has either 1 if a column B is the maximum in a group of values of column A or 0 if it is not.
Example:
[Input]
A B
1 2
2 3
1 4
2 5
[Output]
A B is_max
1 2 0
2 5 0
1 4 1
2 3 0
What I'm trying:
df['is_max'] = 0
df.loc[df.reset_index().groupby('A')['B'].idxmax(),'is_max'] = 1

Fix your code by remove the reset_index
df['is_max'] = 0
df.loc[df.groupby('A')['B'].idxmax(),'is_max'] = 1
df
Out[39]:
A B is_max
0 1 2 0
1 2 3 0
2 1 4 1
3 2 5 1

I make assumption A is your group now that you did not state
df['is_max']=(df['B']==df.groupby('A')['B'].transform('max')).astype(int)
or
df1.groupby('A')['B'].apply(lambda x: x==x.max()).astype(int)

Concatenate groups of multiple dataframes

I have a df1:
a b c
1 0 1 4
2 0 2 5
3 1 1 3
and a second df2:
a b c
1 0 1 5
2 0 2 5
3 1 1 4
These df's have the same goups in a and b. Within groupby of 'a' and 'b' I want df2 underneath df1:
a b c
1 0 1 4
2 0 1 5
3 0 2 5
4 0 2 5
5 1 1 3
6 1 1 4
How can I combine groupby() and concat() to get the desired output?

You can do concat then sort_values
df=pd.concat[df1,df2]).sort_values(['a','b']).reset_index(drop=True)

Creating a column that assigns max value of set of rows by condition to all rows in that group

I have a dataframe that looks like this:
data metadata
A 0
A 1
A 2
A 3
A 4
B 0
B 1
B 2
A 0
A 1
B 0
A 0
A 1
B 0
df.data contains two different categories, A and B. df.metadata stores a running count the number of times a category appears consecutively before the category changes. I want to create a column consecutive_count that assigns the max value of metadata per consecutive group to every row in that group. It should look like this:
data metadata consecutive_count
A 0 4
A 1 4
A 2 4
A 3 4
A 4 4
B 0 2
B 1 2
B 2 2
A 0 1
A 1 1
B 0 0
A 0 1
A 1 1
B 0 0
Please advise. Thank you.

Method 1:
You may try transform max on groupby of each group of data
s = df.data.ne(df.data.shift()).cumsum()
df['consecutive_count'] = df.groupby(s).metadata.transform('max')
Out[96]:
data metadata consecutive_count
0 A 0 4
1 A 1 4
2 A 2 4
3 A 3 4
4 A 4 4
5 B 0 2
6 B 1 2
7 B 2 2
8 A 0 1
9 A 1 1
10 B 0 0
11 A 0 1
12 A 1 1
13 B 0 0
Method 2:
Since metadata is sorted per group, you may reverse dataframe and do groupby cummax
s = df.data.ne(df.data.shift()).cumsum()
df['consecutive_count'] = df[::-1].groupby(s).metadata.cummax()
Out[101]:
data metadata consecutive_count
0 A 0 4
1 A 1 4
2 A 2 4
3 A 3 4
4 A 4 4
5 B 0 2
6 B 1 2
7 B 2 2
8 A 0 1
9 A 1 1
10 B 0 0
11 A 0 1
12 A 1 1
13 B 0 0

Counting Precedant Entries of a column and creating a new varaible of these counts

I have a data frame and I want to count the number of consecutive entries of one column and record the counts in a separate variable. Here is an example:
ID Class
1 A
1 A
2 A
1 B
1 B
1 B
2 B
1 C
1 C
2 A
2 A
2 A
I want in each group ID to count the number of consecutive classes, so the output would look like this:
ID Class Counts
1 A 0
1 A 1
2 A 0
1 B 0
1 B 1
1 B 2
2 B 0
1 C 0
1 C 1
2 A 0
2 A 1
2 A 2
I am not looking the frequency of occurrence of a specific entries like here, rather the consecutive occurrences of an entry on the ID level

You can use cumcount by Series which is create by cumsum of shifted concanecate values by shift:
#use separator which is not in data like _ or ¥
s = df['ID'].astype(str) + '¥' + df['Class']
df['Counts'] = df.groupby(s.ne(s.shift()).cumsum()).cumcount()
print (df)
ID Class Counts
0 1 A 0
1 1 A 1
2 2 A 0
3 1 B 0
4 1 B 1
5 1 B 2
6 2 B 0
7 1 C 0
8 1 C 1
9 2 A 0
10 2 A 1
11 2 A 2
Another solution with ngroup (pandas 0.20.2+):
s = df.groupby(['ID','Class']).ngroup()
df['Counts'] = df.groupby(s.ne(s.shift()).cumsum()).cumcount()
print (df)
ID Class Counts
0 1 A 0
1 1 A 1
2 2 A 0
3 1 B 0
4 1 B 1
5 1 B 2
6 2 B 0
7 1 C 0
8 1 C 1
9 2 A 0
10 2 A 1
11 2 A 2

cumulative number of unique elements for pandas dataframe

i have a pandas data frame
id tag
1 A
1 A
1 B
1 C
1 A
2 B
2 C
2 B
I want to add a column which computes the cumulative number of unique tags over at id level. More specifically, I would like to have
id tag count
1 A 1
1 A 1
1 B 2
1 C 3
1 A 3
2 B 1
2 C 2
2 B 2
For a given id, count will be non-decreasing. Thanks for your help!

I think this does what you want:
unique_count = df.drop_duplicates().groupby('id').cumcount() + 1
unique_count.reindex(df.index).ffill()
The +1 is because the count starts at zero. This only works if the dataframe is sorted by id. Was that intended? You can always sort beforehand.

You can find some other approaches in R and Python here
df = pd.DataFrame({'id':[1,1,1,1,1,2,2,2],'tag':["A","A", "B","C","A","B","C","B"]})
df['count']=df.groupby('id')['tag'].apply(lambda x: (~pd.Series(x).duplicated()).cumsum())
id tag count
0 1 A 1
1 1 A 1
2 1 B 2
3 1 C 3
4 1 A 3
5 2 B 1
6 2 C 2
7 2 B 2

How about this:
d['X'] = 1
d.groupby("Col").X.cumsum()

idt=[1,1,1,1,1,2,2,2]
tag=['A','A','B','C','A','B','C','B']
df=pd.DataFrame(tag,index=idt,columns=['tag'])
df=df.reset_index()
print(df)
index tag
0 1 A
1 1 A
2 1 B
3 1 C
4 1 A
5 2 B
6 2 C
7 2 B
df['uCnt']=df.groupby(['index','tag']).cumcount()+1
print(df)
index tag uCnt
0 1 A 1
1 1 A 2
2 1 B 1
3 1 C 1
4 1 A 3
5 2 B 1
6 2 C 1
7 2 B 2
df['uCnt']=df['uCnt']//df['uCnt']**2
print(df)
index tag uCnt
0 1 A 1
1 1 A 0
2 1 B 1
3 1 C 1
4 1 A 0
5 2 B 1
6 2 C 1
7 2 B 0
df['uCnt']=df.groupby(['index'])['uCnt'].cumsum()
print(df)
index tag uCnt
0 1 A 1
1 1 A 1
2 1 B 2
3 1 C 3
4 1 A 3
5 2 B 1
6 2 C 2
7 2 B 2
df=df.set_index('index')
print(df)
tag uCnt
index
1 A 1
1 A 1
1 B 2
1 C 3
1 A 3
2 B 1
2 C 2
2 B 2

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