Suppose I have two Pandas DataFrames:
df1
with columns k1
( str
), k2
( str
), and v
( float
), and df2
with a column w
( float
). I can assume that the rows df1
are sorted, first by k1
, then by k2
, and finally by v
. I can assume that the values of w
in df2
are unique and sorted.
My goal is to create a new DataFrame df3
with columns k1
, k2
, w
, and count_ge
. The DataFrame df3
should have one row for each unique combination of k1
, k2
, and w
; the column count_ge
should be the number of rows in df1
that have the same values of k1
and k2
, and a value of v
that is greater than or equal to the value of w
.
The following code is a naive implementation that seems to do what I want. Is there an efficient way to carry out the same operation? Ideally, the code should also generalize to more than two keys in df1
.
import pandas as pd
# Generate some example data.
df1 = pd.DataFrame(
(
('A', 'A', 1),
('A', 'A', 1),
('A', 'A', 3),
('A', 'A', 4),
('B', 'C', 2),
('B', 'C', 6),
),
columns=('k1', 'k2', 'v'),
)
df2 = pd.DataFrame(
(0, 2, 5),
columns=('w',),
)
# Get all unique combinations of k1, k2, and w.
# In Pandas 1.2.0, we can use `merge(how='cross')` for this instead.
df3 = (
df1[['k1', 'k2']]
.drop_duplicates()
.assign(_key=1)
.merge(df2.assign(_key=1), on='_key')
.drop(columns='_key')
)
# For each row in df3, count the number of rows in df1 that have the same values of k1 and k2,
# and a value of v that is greater than or equal to w.
df3['count_ge'] = 0
for i, (k1, k2, w, _) in df3.iterrows():
df3.loc[i, 'count_ge'] = len(df1.query(f'k1 == {k1!r} and k2 == {k2!r} and v >= {w!r}'))
df3
Initialize df3
using a cross merge
:
df3 = df1[["k1", "k2"]].drop_duplicates().merge(df2, how='cross')
>>> df3
k1 k2 w
0 A A 0
1 A A 2
2 A A 5
3 B C 0
4 B C 2
5 B C 5
Then for the count_ge
column, you could use a lambda
function like so:
df3['count_ge'] = df3.apply(lambda x: df1[(df1["k1"]==x["k1"])&(df1["k2"]==x["k2"])&(df1["v"]>=x["w"])].shape[0], axis=1)
>>> df3
k1 k2 w count_ge
0 A A 0 4
1 A A 2 2
2 A A 5 0
3 B C 0 2
4 B C 2 2
5 B C 5 1
Another possible approach is to use np.histogram
. This method seems fairly clean, but has the potential drawback of copying the DataFrames in pd.concat
. Other suggestions are still welcome.
import numpy as np
import pandas as pd
# Generate some example data.
df1 = pd.DataFrame(
(
('A', 'A', 1),
('A', 'A', 1),
('A', 'A', 3),
('A', 'A', 4),
('B', 'C', 2),
('B', 'C', 6),
),
columns=('k1', 'k2', 'v'),
)
df2 = pd.DataFrame(
(0, 2, 5),
columns=('w',),
)
# For each unique combination of (k1, k2, w), count the number of rows in df1 that have the same values of k1 and k2,
# and a value of v that is greater than or equal to w.
# such
v_bins = np.concatenate((df2['w'], [np.inf]))
df3s = []
for (k1, k2), v in df1.groupby(['k1', 'k2'])['v']:
df = df2.copy()
df['count_ge'] = np.histogram(a=v, bins=v_bins)[0][::-1].cumsum()[::-1]
df['k1'] = k1
df['k2'] = k2
df3s.append(df[['k1', 'k2', 'w', 'count_ge']])
pd.concat(df3s)
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.