I have a data frame with three string columns. I know that the only one value in the 3rd column is valid for every combination of the first two. To clean the data I have to group by data frame by first two columns and select most common value of the third column for each combination.
My code:
import pandas as pd
from scipy import stats
source = pd.DataFrame({'Country' : ['USA', 'USA', 'Russia','USA'],
'City' : ['New-York', 'New-York', 'Sankt-Petersburg', 'New-York'],
'Short name' : ['NY','New','Spb','NY']})
print source.groupby(['Country','City']).agg(lambda x: stats.mode(x['Short name'])[0])
Last line of code doesn't work, it says "Key error 'Short name'" and if I try to group only by City, then I got an AssertionError. What can I do fix it?
This question is related to
python
pandas
group-by
pandas-groupby
mode
A little late to the game here, but I was running into some performance issues with HYRY's solution, so I had to come up with another one.
It works by finding the frequency of each key-value, and then, for each key, only keeping the value that appears with it most often.
There's also an additional solution that supports multiple modes.
On a scale test that's representative of the data I'm working with, this reduced runtime from 37.4s to 0.5s!
Here's the code for the solution, some example usage, and the scale test:
import numpy as np
import pandas as pd
import random
import time
test_input = pd.DataFrame(columns=[ 'key', 'value'],
data= [[ 1, 'A' ],
[ 1, 'B' ],
[ 1, 'B' ],
[ 1, np.nan ],
[ 2, np.nan ],
[ 3, 'C' ],
[ 3, 'C' ],
[ 3, 'D' ],
[ 3, 'D' ]])
def mode(df, key_cols, value_col, count_col):
'''
Pandas does not provide a `mode` aggregation function
for its `GroupBy` objects. This function is meant to fill
that gap, though the semantics are not exactly the same.
The input is a DataFrame with the columns `key_cols`
that you would like to group on, and the column
`value_col` for which you would like to obtain the mode.
The output is a DataFrame with a record per group that has at least one mode
(null values are not counted). The `key_cols` are included as columns, `value_col`
contains a mode (ties are broken arbitrarily and deterministically) for each
group, and `count_col` indicates how many times each mode appeared in its group.
'''
return df.groupby(key_cols + [value_col]).size() \
.to_frame(count_col).reset_index() \
.sort_values(count_col, ascending=False) \
.drop_duplicates(subset=key_cols)
def modes(df, key_cols, value_col, count_col):
'''
Pandas does not provide a `mode` aggregation function
for its `GroupBy` objects. This function is meant to fill
that gap, though the semantics are not exactly the same.
The input is a DataFrame with the columns `key_cols`
that you would like to group on, and the column
`value_col` for which you would like to obtain the modes.
The output is a DataFrame with a record per group that has at least
one mode (null values are not counted). The `key_cols` are included as
columns, `value_col` contains lists indicating the modes for each group,
and `count_col` indicates how many times each mode appeared in its group.
'''
return df.groupby(key_cols + [value_col]).size() \
.to_frame(count_col).reset_index() \
.groupby(key_cols + [count_col])[value_col].unique() \
.to_frame().reset_index() \
.sort_values(count_col, ascending=False) \
.drop_duplicates(subset=key_cols)
print test_input
print mode(test_input, ['key'], 'value', 'count')
print modes(test_input, ['key'], 'value', 'count')
scale_test_data = [[random.randint(1, 100000),
str(random.randint(123456789001, 123456789100))] for i in range(1000000)]
scale_test_input = pd.DataFrame(columns=['key', 'value'],
data=scale_test_data)
start = time.time()
mode(scale_test_input, ['key'], 'value', 'count')
print time.time() - start
start = time.time()
modes(scale_test_input, ['key'], 'value', 'count')
print time.time() - start
start = time.time()
scale_test_input.groupby(['key']).agg(lambda x: x.value_counts().index[0])
print time.time() - start
Running this code will print something like:
key value
0 1 A
1 1 B
2 1 B
3 1 NaN
4 2 NaN
5 3 C
6 3 C
7 3 D
8 3 D
key value count
1 1 B 2
2 3 C 2
key count value
1 1 2 [B]
2 3 2 [C, D]
0.489614009857
9.19386196136
37.4375009537
Hope this helps!
pd.Series.mode is available!Use groupby, GroupBy.agg, and apply the pd.Series.mode function to each group:
source.groupby(['Country','City'])['Short name'].agg(pd.Series.mode)
Country City
Russia Sankt-Petersburg Spb
USA New-York NY
Name: Short name, dtype: object
If this is needed as a DataFrame, use
source.groupby(['Country','City'])['Short name'].agg(pd.Series.mode).to_frame()
Short name
Country City
Russia Sankt-Petersburg Spb
USA New-York NY
The useful thing about Series.mode is that it always returns a Series, making it very compatible with agg and apply, especially when reconstructing the groupby output. It is also faster.
# Accepted answer.
%timeit source.groupby(['Country','City']).agg(lambda x:x.value_counts().index[0])
# Proposed in this post.
%timeit source.groupby(['Country','City'])['Short name'].agg(pd.Series.mode)
5.56 ms ± 343 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.76 ms ± 387 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Series.mode also does a good job when there are multiple modes:
source2 = source.append(
pd.Series({'Country': 'USA', 'City': 'New-York', 'Short name': 'New'}),
ignore_index=True)
# Now `source2` has two modes for the
# ("USA", "New-York") group, they are "NY" and "New".
source2
Country City Short name
0 USA New-York NY
1 USA New-York New
2 Russia Sankt-Petersburg Spb
3 USA New-York NY
4 USA New-York New
source2.groupby(['Country','City'])['Short name'].agg(pd.Series.mode)
Country City
Russia Sankt-Petersburg Spb
USA New-York [NY, New]
Name: Short name, dtype: object
Or, if you want a separate row for each mode, you can use GroupBy.apply:
source2.groupby(['Country','City'])['Short name'].apply(pd.Series.mode)
Country City
Russia Sankt-Petersburg 0 Spb
USA New-York 0 NY
1 New
Name: Short name, dtype: object
If you don't care which mode is returned as long as it's either one of them, then you will need a lambda that calls mode and extracts the first result.
source2.groupby(['Country','City'])['Short name'].agg(
lambda x: pd.Series.mode(x)[0])
Country City
Russia Sankt-Petersburg Spb
USA New-York NY
Name: Short name, dtype: object
You can also use statistics.mode from python, but...
source.groupby(['Country','City'])['Short name'].apply(statistics.mode)
Country City
Russia Sankt-Petersburg Spb
USA New-York NY
Name: Short name, dtype: object
...it does not work well when having to deal with multiple modes; a StatisticsError is raised. This is mentioned in the docs:
If data is empty, or if there is not exactly one most common value, StatisticsError is raised.
But you can see for yourself...
statistics.mode([1, 2])
# ---------------------------------------------------------------------------
# StatisticsError Traceback (most recent call last)
# ...
# StatisticsError: no unique mode; found 2 equally common values
Formally, the correct answer is the @eumiro Solution. The problem of @HYRY solution is that when you have a sequence of numbers like [1,2,3,4] the solution is wrong, i. e., you don't have the mode. Example:
>>> import pandas as pd
>>> df = pd.DataFrame(
{
'client': ['A', 'B', 'A', 'B', 'B', 'C', 'A', 'D', 'D', 'E', 'E', 'E', 'E', 'E', 'A'],
'total': [1, 4, 3, 2, 4, 1, 2, 3, 5, 1, 2, 2, 2, 3, 4],
'bla': [10, 40, 30, 20, 40, 10, 20, 30, 50, 10, 20, 20, 20, 30, 40]
}
)
If you compute like @HYRY you obtain:
>>> print(df.groupby(['client']).agg(lambda x: x.value_counts().index[0]))
total bla
client
A 4 30
B 4 40
C 1 10
D 3 30
E 2 20
Which is clearly wrong (see the A value that should be 1 and not 4) because it can't handle with unique values.
Thus, the other solution is correct:
>>> import scipy.stats
>>> print(df.groupby(['client']).agg(lambda x: scipy.stats.mode(x)[0][0]))
total bla
client
A 1 10
B 4 40
C 1 10
D 3 30
E 2 20
A slightly clumsier but faster approach for larger datasets involves getting the counts for a column of interest, sorting the counts highest to lowest, and then de-duplicating on a subset to only retain the largest cases. The code example is following:
>>> import pandas as pd
>>> source = pd.DataFrame(
{
'Country': ['USA', 'USA', 'Russia', 'USA'],
'City': ['New-York', 'New-York', 'Sankt-Petersburg', 'New-York'],
'Short name': ['NY', 'New', 'Spb', 'NY']
}
)
>>> grouped_df = source\
.groupby(['Country','City','Short name'])[['Short name']]\
.count()\
.rename(columns={'Short name':'count'})\
.reset_index()\
.sort_values('count', ascending=False)\
.drop_duplicates(subset=['Country', 'City'])\
.drop('count', axis=1)
>>> print(grouped_df)
Country City Short name
1 USA New-York NY
0 Russia Sankt-Petersburg Spb
The two top answers here suggest:
df.groupby(cols).agg(lambda x:x.value_counts().index[0])
or, preferably
df.groupby(cols).agg(pd.Series.mode)
However both of these fail in simple edge cases, as demonstrated here:
df = pd.DataFrame({
'client_id':['A', 'A', 'A', 'A', 'B', 'B', 'B', 'C'],
'date':['2019-01-01', '2019-01-01', '2019-01-01', '2019-01-01', '2019-01-01', '2019-01-01', '2019-01-01', '2019-01-01'],
'location':['NY', 'NY', 'LA', 'LA', 'DC', 'DC', 'LA', np.NaN]
})
The first:
df.groupby(['client_id', 'date']).agg(lambda x:x.value_counts().index[0])
yields IndexError (because of the empty Series returned by group C). The second:
df.groupby(['client_id', 'date']).agg(pd.Series.mode)
returns ValueError: Function does not reduce, since the first group returns a list of two (since there are two modes). (As documented here, if the first group returned a single mode this would work!)
Two possible solutions for this case are:
import scipy
x.groupby(['client_id', 'date']).agg(lambda x: scipy.stats.mode(x)[0])
And the solution given to me by cs95 in the comments here:
def foo(x):
m = pd.Series.mode(x);
return m.values[0] if not m.empty else np.nan
df.groupby(['client_id', 'date']).agg(foo)
However, all of these are slow and not suited for large datasets. A solution I ended up using which a) can deal with these cases and b) is much, much faster, is a lightly modified version of abw33's answer (which should be higher):
def get_mode_per_column(dataframe, group_cols, col):
return (dataframe.fillna(-1) # NaN placeholder to keep group
.groupby(group_cols + [col])
.size()
.to_frame('count')
.reset_index()
.sort_values('count', ascending=False)
.drop_duplicates(subset=group_cols)
.drop(columns=['count'])
.sort_values(group_cols)
.replace(-1, np.NaN)) # restore NaNs
group_cols = ['client_id', 'date']
non_grp_cols = list(set(df).difference(group_cols))
output_df = get_mode_per_column(df, group_cols, non_grp_cols[0]).set_index(group_cols)
for col in non_grp_cols[1:]:
output_df[col] = get_mode_per_column(df, group_cols, col)[col].values
Essentially, the method works on one col at a time and outputs a df, so instead of concat, which is intensive, you treat the first as a df, and then iteratively add the output array (values.flatten()) as a column in the df.
If you want another approach for solving it that is does not depend on value_counts or scipy.stats you can use the Counter collection
from collections import Counter
get_most_common = lambda values: max(Counter(values).items(), key = lambda x: x[1])[0]
Which can be applied to the above example like this
src = pd.DataFrame({'Country' : ['USA', 'USA', 'Russia','USA'],
'City' : ['New-York', 'New-York', 'Sankt-Petersburg', 'New-York'],
'Short_name' : ['NY','New','Spb','NY']})
src.groupby(['Country','City']).agg(get_most_common)
If you don't want to include NaN values, using Counter is much much faster than pd.Series.mode or pd.Series.value_counts()[0]:
def get_most_common(srs):
x = list(srs)
my_counter = Counter(x)
return my_counter.most_common(1)[0][0]
df.groupby(col).agg(get_most_common)
should work. This will fail when you have NaN values, as each NaN will be counted separately.
The problem here is the performance, if you have a lot of rows it will be a problem.
If it is your case, please try with this:
import pandas as pd
source = pd.DataFrame({'Country' : ['USA', 'USA', 'Russia','USA'],
'City' : ['New-York', 'New-York', 'Sankt-Petersburg', 'New-York'],
'Short_name' : ['NY','New','Spb','NY']})
source.groupby(['Country','City']).agg(lambda x:x.value_counts().index[0])
source.groupby(['Country','City']).Short_name.value_counts().groupby['Country','City']).first()
For agg, the lambba function gets a Series, which does not have a 'Short name' attribute.
stats.mode returns a tuple of two arrays, so you have to take the first element of the first array in this tuple.
With these two simple changements:
source.groupby(['Country','City']).agg(lambda x: stats.mode(x)[0][0])
returns
Short name
Country City
Russia Sankt-Petersburg Spb
USA New-York NY
Source: Stackoverflow.com