I have a df:
id value
1 10
2 15
1 10
1 10
2 13
3 10
3 20
I am trying to keep only rows that have 1 unique value in column value so that the result df looks like this:
id value
1 10
1 10
1 10
I dropped id = 2, 3 because it has more than 1 unique value in column value, 15, 13 & 10, 20 respectively.
I read this answer.
But this simply removes duplicates whereas I want to check if a given column - in this case column value has more than 1 unique value.
I tried:
df['uniques'] = pd.Series(df.groupby('id')['value'].nunique())
But this returns nan for every row since I am trying to fit n returns on n+m rows after grouping. I can write a function and apply it to every row but I was wondering if there is a smart quick filter that achieves my goal.
Use transform with groupby to align the group values to the individual rows:
df['nuniques'] = df.groupby('id')['value'].transform('nunique')
Output:
id value nuniques
0 1 10 1
1 2 15 2
2 1 10 1
3 1 10 1
4 2 13 2
5 3 10 2
6 3 20 2
If you only need to filter your data, you don't need to assign the new column:
df[df.groupby('id')['value'].transform('nunique') == 1]
Let us do filter
out = df.groupby('id').filter(lambda x : x['value'].nunique()==1)
Out[6]:
id value
0 1 10
2 1 10
3 1 10
I have a dataframe like this:
user_id order_id
0 a 1
1 a 2
2 a 3
3 b 4
4 c 5
Now I want to add a column to show whether the user of each order has multiple orders:
user_id order_id repetitive
0 a 1 1
1 a 2 1
2 a 3 1
3 b 4 0
4 c 5 0
Since a has three orders, the tag is 1. I know the method value_counts can calculate the result but it only shows the result after groupby. I want to combine it with the origin dataframe. How can I achieve this?
Use groupby and transform to get your counts while maintaining the same structure.
df['repetitive'] = df.groupby('user_id').transform('count').gt(1).astype(int)
Suppose I have pandas DataFrame like this:
df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4], 'value':[1,2,3,1,2,3,4,1,1]})
which looks like:
id value
0 1 1
1 1 2
2 1 3
3 2 1
4 2 2
5 2 3
6 2 4
7 3 1
8 4 1
I want to get a new DataFrame with top 2 records for each id, like this:
id value
0 1 1
1 1 2
3 2 1
4 2 2
7 3 1
8 4 1
I can do it with numbering records within group after groupby:
dfN = df.groupby('id').apply(lambda x:x['value'].reset_index()).reset_index()
which looks like:
id level_1 index value
0 1 0 0 1
1 1 1 1 2
2 1 2 2 3
3 2 0 3 1
4 2 1 4 2
5 2 2 5 3
6 2 3 6 4
7 3 0 7 1
8 4 0 8 1
then for the desired output:
dfN[dfN['level_1'] <= 1][['id', 'value']]
Output:
id value
0 1 1
1 1 2
3 2 1
4 2 2
7 3 1
8 4 1
But is there more effective/elegant approach to do this? And also is there more elegant approach to number records within each group (like SQL window function row_number()).
Did you try
df.groupby('id').head(2)
Output generated:
id value
id
1 0 1 1
1 1 2
2 3 2 1
4 2 2
3 7 3 1
4 8 4 1
(Keep in mind that you might need to order/sort before, depending on your data)
EDIT: As mentioned by the questioner, use
df.groupby('id').head(2).reset_index(drop=True)
to remove the MultiIndex and flatten the results:
id value
0 1 1
1 1 2
2 2 1
3 2 2
4 3 1
5 4 1
Since 0.14.1, you can now do nlargest and nsmallest on a groupby object:
In [23]: df.groupby('id')['value'].nlargest(2)
Out[23]:
id
1 2 3
1 2
2 6 4
5 3
3 7 1
4 8 1
dtype: int64
There's a slight weirdness that you get the original index in there as well, but this might be really useful depending on what your original index was.
If you're not interested in it, you can do .reset_index(level=1, drop=True) to get rid of it altogether.
(Note: From 0.17.1 you'll be able to do this on a DataFrameGroupBy too but for now it only works with Series and SeriesGroupBy.)
Sometimes sorting the whole data ahead is very time consuming.
We can groupby first and doing topk for each group:
g = df.groupby(['id']).apply(lambda x: x.nlargest(topk,['value'])).reset_index(drop=True)
df.groupby('id').apply(lambda x : x.sort_values(by = 'value', ascending = False).head(2).reset_index(drop = True))
Here sort values ascending false gives similar to nlargest and True gives similar to nsmallest.
The value inside the head is the same as the value we give inside nlargest to get the number of values to display for each group.
reset_index is optional and not necessary.
This works for duplicated values
If you have duplicated values in top-n values, and want only unique values, you can do like this:
import pandas as pd
ifile = "https://raw.githubusercontent.com/bhishanpdl/Shared/master/data/twitter_employee.tsv"
df = pd.read_csv(ifile,delimiter='\t')
print(df.query("department == 'Audit'")[['id','first_name','last_name','department','salary']])
id first_name last_name department salary
24 12 Shandler Bing Audit 110000
25 14 Jason Tom Audit 100000
26 16 Celine Anston Audit 100000
27 15 Michale Jackson Audit 70000
If we do not remove duplicates, for the audit department we get top 3 salaries as 110k,100k and 100k.
If we want to have not-duplicated salaries per each department, we can do this:
(df.groupby('department')['salary']
.apply(lambda ser: ser.drop_duplicates().nlargest(3))
.droplevel(level=1)
.sort_index()
.reset_index()
)
This gives
department salary
0 Audit 110000
1 Audit 100000
2 Audit 70000
3 Management 250000
4 Management 200000
5 Management 150000
6 Sales 220000
7 Sales 200000
8 Sales 150000
To get the first N rows of each group, another way is via groupby().nth[:N]. The outcome of this call is the same as groupby().head(N). For example, for the top-2 rows for each id, call:
N = 2
df1 = df.groupby('id', as_index=False).nth[:N]
To get the largest N values of each group, I suggest two approaches.
First sort by "id" and "value" (make sure to sort "id" in ascending order and "value" in descending order by using the ascending parameter appropriately) and then call groupby().nth[].
N = 2
df1 = df.sort_values(by=['id', 'value'], ascending=[True, False])
df1 = df1.groupby('id', as_index=False).nth[:N]
Another approach is to rank the values of each group and filter using these ranks.
# for the entire rows
N = 2
msk = df.groupby('id')['value'].rank(method='first', ascending=False) <= N
df1 = df[msk]
# for specific column rows
df1 = df.loc[msk, 'value']
Both of these are much faster than groupby().apply() and groupby().nlargest() calls as suggested in the other answers on here(1, 2, 3). On a sample with 100k rows and 8000 groups, a %timeit test showed that it was 24-150 times faster than those solutions.
Also, instead of slicing, you can also pass a list/tuple/range to a .nth() call:
df.groupby('id', as_index=False).nth([0,1])
# doesn't even have to be consecutive
# the following returns 1st and 3rd row of each id
df.groupby('id', as_index=False).nth([0,2])
I have a dataframe like this:
userid itemid timestamp
1 1 50
1 2 50
1 3 50
1 4 60
2 1 40
2 2 50
I want to drop all rows whose userid occur more than 2 times and get a new dataframe as follows. Does someone can help me? Thanks.
userid itemid timestamp
2 1 40
2 2 50
You can use pd.Series.value_counts and calculate an array of userid filtered by your condition. Then use this to filter your original dataframe.
c = df['userid'].value_counts()
idx = c[c > 2].index
res = df[~df['userid'].isin(idx)]
print(res)
userid itemid timestamp
4 2 1 40
5 2 2 50
I'm trying to generate a timeseries from a dataframe, but the solutions I've found here don't really address my specific problem. I have a dataframe which is a series of id's which iterate from 1 to n, then repeat, like this:
key ID Var_1
0 1 1
0 2 1
0 3 2
1 1 3
1 2 2
1 3 1
I want to reshape it into a timeseries in which the index
ID Var_1_0 Var_2_0
1 1 3
2 1 2
3 2 1
I have tried the stack() method but it doesn't generate the result I want. Generating an index from ID seems to be the right ID is not a proper date so I'm not sure how to proceed. Pointers much appreciated.
Try this:
import pandas as pd
df = pd.DataFrame([[0,1,1], [0,2,1], [0,3,2], [1,1,3], [1,2,2], [1,3,1]], columns=('key', 'ID', 'Var_1'))
Use the pivot function:
df2 = df.pivot('ID', 'key', 'Var_1')
You can rename the columns by:
df2.columns = ('Var_1_0', 'Var_2_0')
Result:
Out:
Var_1_0 Var_2_0
ID
1 1 3
2 1 2
3 2 1