I am trying to make segregate my data into buckets based on certain user attributes and I would like to see some counts in each of the buckets.For this I have imported this data into a Pandas Dataframe.
I have data that has user city, kids age and their unique id. I would like to know the count of users who reside in city A and have kids in age group 0-5.
Sample Data frame looks something like this:
city kids_age user_id
A 10 1
B 4 2
A 4 3
C 8 4
A 3 5
Expected Output:
city bin count
A 0-5 2
5-10 1
B 0-5 1
5-10 0
C 0-5 0
5-10 1
I tried group by on two columns city and kids age:
user_details_df_cropped_1.groupby(['city', 'kids_age']).count()
It gave me an output that looks something like this:
city kids_age user_id count
A 10 1 1
4 3 1
3 5 1
B 4 2 1
C 8 4 1
I returns me the users grouped by city, but not really by kids age bins(ranges). What am I missing here? Appreciate the help!!
Use cut for binning, pass to DataFrame.groupby, add 0 rows with DataFrame.stack
DataFrame.unstack an last convert to DataFrame by Series.reset_index:
bins = [0,5,10]
labels = ['{}-{}'.format(i, j) for i, j in zip(bins[:-1], bins[1:])]
b = pd.cut(df['kids_age'], bins=bins, labels=labels, include_lowest=True)
df = df.groupby(['city', b]).size().unstack(fill_value=0).stack().reset_index(name='count')
print (df)
city kids_age count
0 A 0-5 2
1 A 5-10 1
2 B 0-5 1
3 B 5-10 0
4 C 0-5 0
5 C 5-10 1
Another solution with DataFrame.reindex and MultiIndex.from_product for added mising rows filled by 0:
bins = [0,5,10]
labels = ['{}-{}'.format(i, j) for i, j in zip(bins[:-1], bins[1:])]
b = pd.cut(df['kids_age'], bins=bins, labels=labels, include_lowest=True)
mux = pd.MultiIndex.from_product([df['city'].unique(), labels], names=['city','kids_age'])
df = (df.groupby(['city', b])
.size()
.reindex(mux, fill_value=0)
.reset_index(name='count'))
print (df)
city kids_age count
0 A 0-5 2
1 A 5-10 1
2 B 0-5 1
3 B 5-10 0
4 C 0-5 0
5 C 5-10 1
Related
A conflict row is that two rows have same feature but with different label, like this:
feature label
a 1
a 0
Now, I want to merge these conflict rows to only one label getting from their counts. If I have more a 1, then a will be labeled as 1. Otherwise, a should be labeled as 0.
I can find these conflicts by df1=df.groupy('feature', as_index=Fasle).nunique(),df1 = df1[df1['label]==2]' , and their value counts by df2 = df.groupby("feature")["label"].value_counts().reset_index(name="counts").
But how to find these conflic rows and their counts in one Dataframe (df_conflict = ?), and then merge them by counts, (df_merged = merge(df))?
Lets take df = pd.DataFrame({"feature":['a','a','b','b','a','c','c','d'],'label':[1,0,0,1,1,0,0,1]}) as example.
feature label
0 a 1
1 a 0
2 b 0
3 b 1
4 a 1
5 c 0
6 c 0
7 d 1
df_conflict should be :
feature label counts
a 1 2
a 0 1
b 0 1
b 1 1
And df_merged will be:
feature label
a 1
b 0
c 0
d 1
I think you need first filter groups with count of unique values by DataFrameGroupBy.nunique with GroupBy.transform before SeriesGroupBy.value_counts:
df1 = df[df.groupby('feature')['label'].transform('nunique').gt(1)]
df_conflict = df1.groupby('feature')['label'].value_counts().reset_index(name='count')
print (df_conflict)
feature label count
0 a 1 2
1 a 0 1
2 b 0 1
3 b 1 1
For second get feature with labels by maximum occurencies:
df_merged = df.groupby('feature')['label'].agg(lambda x: x.value_counts().index[0]).reset_index()
print (df_merged)
feature label
0 a 1
1 b 0
2 c 0
3 d 1
This is my pandas df:
Id Protein A_Egg B_Meat C_Milk Category
A 10 10 20 0 egg
B 20 10 0 10 milk
C 20 10 10 10 meat
D 25 20 10 0 egg
I wish to merge protein column with other column based on "Category"
My output is
Id Protein_final
A 20
B 30
C 30
D 45
Ideally, I would like to show how I am approaching but, I am frankly clueless!!
EDIT: Also, How to handle is the category is blank or does meet one of the column (in that can final should be same as initial value in protein column)
Use DataFrame.lookup with some preprocessing with remove values in columns names before _ and lowercase, last add to column:
arr = df.rename(columns=lambda x: x.split('_')[-1].lower()).lookup(df.index, df['Category'])
df['Protein'] += arr
print (df)
Id Protein A_Egg B_Meat C_Milk Category
0 A 20 10 20 0 egg
1 B 30 10 0 10 milk
2 C 30 10 10 10 meat
3 D 45 20 10 0 egg
If need only 2 columns finally:
df = df[['Id','Protein']]
You can melt the dataframe, and filter for rows where category equals the variable column, and sum the final columns :
(
df
.melt(["Id", "Protein", "Category"])
.assign(variable=lambda x: x.variable.str[2:].str.lower(),
Protein_final=lambda x: x.Protein + x.value)
.query("Category == variable")
.filter(["Id", "Protein_final"])
)
Id Protein_final
0 A 20
3 D 45
6 C 30
9 B 30
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
I want to create new columns out of the unique values of one column with the count of the unique values as values assigned in the row.
df = pd.DataFrame([["a",20],["a", 10],["b", 5],["c",10],
["b", 10],["a", 5],["c",5],["c",5]],
columns=["alp","min"])
In [4]: df
Out[4]:
alp min
0 a 20
1 a 10
2 b 5
3 c 10
4 b 10
5 a 5
6 c 5
7 c 5
I tried using groupby to get the values I want
In [8]: df.groupby('alp')['min'].count()
Out[8]:
alp
a 3
b 2
c 3
Name: min, dtype: int64
Now, I want to create columns out of that output.
count_a count_b count_c
0 3 2 3
Is there any function to achieve this in pandas?
Remove Series name by Series.rename_axis, convert to one column DataFrame by Series.to_frame, transpose by DataFrame.T and last DataFrame.add_prefix:
df = df.groupby('alp')['min'].count().rename_axis(None).to_frame(0).T.add_prefix('count_')
print (df)
count_a count_b count_c
0 3 2 3
Or create DataFrame per constructor:
s = df.groupby('alp')['min'].count()
df = pd.DataFrame([s.values], columns='count_' + s.index.values)
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])