I have the following Pandas DataFrame:
ID CAT
1 A
1 B
1 A
2 A
2 B
2 A
1 B
1 A
I'd like to have a table that indicates the number of occurance per CAT values for each ID in different columns like this:
ID CAT_A_NUM CAT_B_NUM
1 3 2
2 2 1
I tried in many ways, like this one with pivot table, but unsuccessfully:
df.pivot_table(values='CAT', index='ID', columns='CAT', aggfunc='count')
you can use crosstab():
df=pd.DataFrame(data={'ID':[1,1,1,2,2,2,1,1],'CAT':['A','B','A','A','B','A','B','A']})
final = pd.crosstab(df['ID'], df['CAT'])
final.columns=['CAT_A_NUM','CAT_B_NUM']
final
ID CAT_A_NUM CAT_B_NUM
1 3 2
2 2 1
Probably you can use groupby + unstack
df.groupby(["ID","CAT"]).size().unstack()
which gives
CAT A B
ID
1 3 2
2 2 1
I have a df that looks like below:
ID Name Supervisor SupervisorID
1 X Y
2 Y C
3 Z Y
4 C Y
5 V X
What I need is to find SupervisorID. I can find his ID by checking it in column Name and that I will see his ID so if supervisor is Y then I see that in column Name there is Y so his ID id 2. DF should looks like below:
ID Name Supervisor SupervisorID
1 X Y 2
2 Y C 4
3 Z Y 2
4 C Y 2
5 V X 1
Do you have any idea how to solve this?
Thanks for help and best regards
Use Series.map with DataFrame.drop_duplicates for unique Names, because in real data duplicates:
df['SupervisorID']=df['Supervisor'].map(df.drop_duplicates('Name').set_index('Name')['ID'])
print (df)
ID Name Supervisor SupervisorID
0 1 X Y 2
1 2 Y C 4
2 3 Z Y 2
3 4 C Y 2
4 5 V X 1
I am removing consecutive duplicates in groups in a dataframe. I am looking for a faster way than this:
def remove_consecutive_dupes(subdf):
dupe_ids = [ "A", "B" ]
is_duped = (subdf[dupe_ids].shift(-1) == subdf[dupe_ids]).all(axis=1)
subdf = subdf[~is_duped]
return subdf
# dataframe with columns key, A, B
df.groupby("key").apply(remove_consecutive_dupes).reset_index()
Is it possible to remove these without grouping first? Applying the above function to each group individually takes a lot of time, especially if the group count is like half the row count. Is there a way to do this operation on the entire dataframe at once?
A simple example for the algorithm if the above was not clear:
input:
key A B
0 x 1 2
1 y 1 4
2 x 1 2
3 x 1 4
4 y 2 5
5 x 1 2
output:
key A B
0 x 1 2
1 y 1 4
3 x 1 4
4 y 2 5
5 x 1 2
Row 2 was dropped because A=1 B=2 was also the previous row in group x.
Row 5 will not be dropped because it is not a consecutive duplicate in group x.
According to your code, you drop only lines if they appear below each other if
they are grouped by the key. So rows with another key inbetween do not influence this logic. But doing this, you want to preserve the original order of the records.
I guess the biggest influence in the runtime is the call of your function and
possibly not the grouping itself.
If you want to avoid this, you can try the following approach:
# create a column to restore the original order of the dataframe
df.reset_index(drop=True, inplace=True)
df.reset_index(drop=False, inplace=True)
df.columns= ['original_order'] + list(df.columns[1:])
# add a group column, that contains consecutive numbers if
# two consecutive rows differ in at least one of the columns
# key, A, B
compare_columns= ['key', 'A', 'B']
df.sort_values(['key', 'original_order'], inplace=True)
df['group']= (df[compare_columns] != df[compare_columns].shift(1)).any(axis=1).cumsum()
df.drop_duplicates(['group'], keep='first', inplace=True)
df.drop(columns=['group'], inplace=True)
# now just restore the original index and it's order
df.set_index('original_order', inplace=True)
df.sort_index(inplace=True)
df
Testing this, results in:
key A B
original_order
0 x 1 2
1 y 1 4
3 x 1 4
4 y 2 5
If you don't like the index name above (original_order), you just need to add the following line to remove it:
df.index.name= None
Testdata:
from io import StringIO
infile= StringIO(
""" key A B
0 x 1 2
1 y 1 4
2 x 1 2
3 x 1 4
4 y 2 5"""
)
df= pd.read_csv(infile, sep='\s+') #.set_index('Date')
df
I have a data frame in pandas like this:
STATUS FEATURES
A [x,y,z]
A [t, y]
B [x,p,t]
B [x,p]
I want to count the frequency of the elements in the lists of features conditional on the status.
The desired output would be:
STATUS FEATURES FREQUENCY
A x 1
A y 2
A z 1
A t 1
B x 2
B t 1
B p 2
Let us do explode , the groupby size
s=df.explode(['FEATURES']).groupby(['STATUS','FEATURES']).size().reset_index()
Use DataFrame.explode and SeriesGroupBy.value_counts:
new_df = (df.explode('FEATURES')
.groupby('STATUS')['FEATURES']
.value_counts()
.reset_index(name='FRECUENCY'))
print(new_df)
Output
STATUS FEATURES FRECUENCY
0 A y 2
1 A t 1
2 A x 1
3 A z 1
4 B p 2
5 B x 2
6 B t 1
I have a dataset containing 3 columns, I’m trying to group them and print each group in sorted fashion (based on highest value in each group). The records in each group also have to be in sorted fashion.
Dataset looks like below.
key1,key2,val
b,y,21
c,y,25
c,z,10
b,x,20
b,z,5
c,x,17
a,x,15
a,y,18
a,z,100
df=pd.read_csv('/tmp/hello.csv')
df['max'] = df.groupby(['key1'])['val'].transform('max')
dff=df.sort_values(['max', 'val'], ascending=False).drop('max', axis=1)
I'm applying transform as it works per group basis and then sorting the values.
Above code results in my desired dataframe:
a,z,100
a,y,18
a,x,15
c,y,25
c,x,17
c,z,10
b,y,21
b,x,20
b,z,5
But, the same code fails for below dataset.
key1,key2,val
b,y,10
c,y,10
c,z,10
b,x,2
b,z,2
c,x,2
a,x,2
a,y,2
a,z,2
Below is the desired output
key1,key2,val
c,y,10
c,z,10
c,x,2
b,y,10
b,x,2
b,z,2
a,x,2
a,y,2
a,z,2
Please help me in properly grouping and sorting the dataframe for my scenario.
Add column key1 to sort_values because in second DataFrame are multiple maximum values 10 per groups, so sorting cannot distingush groups:
df['max'] = df.groupby(['key1'])['val'].transform('max')
dff=df.sort_values(['max','key1', 'val'], ascending=False).drop('max', axis=1)
print (dff)
key1 key2 val
8 a z 100
7 a y 18
6 a x 15
1 c y 25
5 c x 17
2 c z 10
0 b y 21
3 b x 20
4 b z 5
df['max'] = df.groupby(['key1'])['val'].transform('max')
dff=df.sort_values(['max','key1', 'val'], ascending=False).drop('max', axis=1)
print (dff)
key1 key2 val
1 c y 10
2 c z 10
5 c x 2
0 b y 10
3 b x 2
4 b z 2
6 a x 2
7 a y 2
8 a z 2