Pandas - Opposite of drop duplicates, keep first

Question

I'm familiar with how to drop duplicate rows, and then using the parameter of first , last , none . Nothing too complicated with that and there's plenty of examples (ie here ).

However, what I'm looking for is there a way to find the duplicates, but instead of dropping all duplicates and keeping the first, if I have duplicates, keep all duplicates but drop the first:

So instead of "drop if duplicates, keep the first", I want "keep if duplicates, drop first"

Example:

Given this dataframe, and looking at duplicates in cost column:

    ID name type cost
0    0    a   bb    1
1    1    a   cc    2 <--- there are duplicates, so drop this row
2  1_0    a   dd    2
3    2    a   ee    3 <--- there are duplicates, so drop this row
4  2_0    a   ff    3
5  2_1    a   gg    3
6  2_2    a   hh    3

If there are duplicates in the cost column, just drop the first occurrence, but keep the rest.

So my output would be:

    ID name type cost
0    0    a   bb    1
2  1_0    a   dd    2
4  2_0    a   ff    3
5  2_1    a   gg    3
6  2_2    a   hh    3

Heres the sample dataframe:

import pandas as pd

df = pd.DataFrame([
['0',   'a',    'bb',   '1'],
['1',   'a',    'cc',   '2'],
['1_0', 'a',    'dd',   '2'],
['2',   'a',    'ee',   '3'],
['2_0', 'a',    'ff',   '3'],
['2_1', 'a',    'gg',   '3'],
['2_2', 'a',    'hh',   '3']], columns = ['ID', 'name', 'type', 'cost'])

Answer 1

You can chain 2 masks created by DataFrame.duplicated with bitwise OR and filter by boolean indexing :

df = df[df.duplicated('cost') | ~df.duplicated('cost', keep=False)]
print (df)
    ID name type cost
0    0    a   bb    1
2  1_0    a   dd    2
4  2_0    a   ff    3
5  2_1    a   gg    3
6  2_2    a   hh    3

Detail :

print (df.assign(mask1=df.duplicated('cost'), mask2=~df.duplicated('cost', keep=False)))
    ID name type cost  mask1  mask2
0    0    a   bb    1  False   True
1    1    a   cc    2  False  False
2  1_0    a   dd    2   True  False
3    2    a   ee    3  False  False
4  2_0    a   ff    3   True  False
5  2_1    a   gg    3   True  False
6  2_2    a   hh    3   True  False

Answer 2

You can do the following with the XOR (^) operator which looks for both conditions to be True. Since we use the NOT (~) operator. It looks for the opposite eg: both False :

df[~(df.cost.duplicated(keep=False) ^ df.cost.duplicated())]

Output

    ID name type cost
0    0    a   bb    1
2  1_0    a   dd    2
4  2_0    a   ff    3
5  2_1    a   gg    3
6  2_2    a   hh    3

Answer 3

You can use groupby and pass a lambda function to grab the records after the first dupe if the dupe exists:

>>> df.groupby('cost').apply(lambda group: group.iloc[1:] if len(group) > 1 else group).reset_index(drop=True)
    ID  cost name type
0    0     1    a   bb
1  1_0     2    a   dd
2  2_0     3    a   ff
3  2_1     3    a   gg
4  2_2     3    a   hh

Answer 4

You can use the following code:

# Import pandas library 
import pandas as pd 

# initialize list of lists so i can create duplicate datas
data = [['tom', 10], ['nick', 15], ['juli', 14], ['nick', 15], ['julia', 140],
        ['tom', 10],['tom', 10],['tom', 10]] 

# Create the pandas DataFrame 
df = pd.DataFrame(data, columns = ['Name', 'Age']) 

# print dataframe. 
print(df)

# Now the logic begins from here

colnames=[]

for col in df.columns:
    colnames.append(col)


listdf=df.values.tolist()
temp=[]

for i in range(0,len(listdf)):
    if(listdf.count(listdf[i])>1 and listdf[i] not in temp):
        temp.append(listdf[i])

df = pd.DataFrame(temp, columns =colnames)

print("dataframe with only duplciates ")
print(df)