问题
For every pair of src and dest airport cities I want to return a percentile of column a given a value of column b.
I can do this manually as such:
example df with only 2 pairs of src/dest (I have thousands in my actual df):
dt src dest a b
0 2016-01-01 YYZ SFO 548.12 279.28
1 2016-01-01 DFW PDX 111.35 -65.50
2 2016-02-01 YYZ SFO 64.84 342.35
3 2016-02-01 DFW PDX 63.81 61.64
4 2016-03-01 YYZ SFO 614.29 262.83
{'a': {0: 548.12,
1: 111.34999999999999,
2: 64.840000000000003,
3: 63.810000000000002,
4: 614.28999999999996,
5: -207.49000000000001,
6: 151.31999999999999,
7: -56.43,
8: 611.37,
9: -296.62,
10: 6417.5699999999997,
11: -376.25999999999999,
12: 465.12,
13: -821.73000000000002,
14: 1270.6700000000001,
15: -1410.0899999999999,
16: 1312.6600000000001,
17: -326.25999999999999,
18: 1683.3699999999999,
19: -24.440000000000001,
20: 583.60000000000002,
21: -5.2400000000000002,
22: 1122.74,
23: 195.21000000000001,
24: 97.040000000000006,
25: 133.94},
'b': {0: 279.27999999999997,
1: -65.5,
2: 342.35000000000002,
3: 61.640000000000001,
4: 262.82999999999998,
5: 115.89,
6: 268.63999999999999,
7: 2.3500000000000001,
8: 91.849999999999994,
9: 62.119999999999997,
10: 778.33000000000004,
11: -142.78,
12: 1675.53,
13: -214.36000000000001,
14: 983.80999999999995,
15: -207.62,
16: 632.13999999999999,
17: -132.53,
18: 422.36000000000001,
19: 13.470000000000001,
20: 642.73000000000002,
21: -144.59999999999999,
22: 213.15000000000001,
23: -50.200000000000003,
24: 338.27999999999997,
25: -129.69},
'dest': {0: 'SFO',
1: 'PDX',
2: 'SFO',
3: 'PDX',
4: 'SFO',
5: 'PDX',
6: 'SFO',
7: 'PDX',
8: 'SFO',
9: 'PDX',
10: 'SFO',
11: 'PDX',
12: 'SFO',
13: 'PDX',
14: 'SFO',
15: 'PDX',
16: 'SFO',
17: 'PDX',
18: 'SFO',
19: 'PDX',
20: 'SFO',
21: 'PDX',
22: 'SFO',
23: 'PDX',
24: 'SFO',
25: 'PDX'},
'dt': {0: Timestamp('2016-01-01 00:00:00'),
1: Timestamp('2016-01-01 00:00:00'),
2: Timestamp('2016-02-01 00:00:00'),
3: Timestamp('2016-02-01 00:00:00'),
4: Timestamp('2016-03-01 00:00:00'),
5: Timestamp('2016-03-01 00:00:00'),
6: Timestamp('2016-04-01 00:00:00'),
7: Timestamp('2016-04-01 00:00:00'),
8: Timestamp('2016-05-01 00:00:00'),
9: Timestamp('2016-05-01 00:00:00'),
10: Timestamp('2016-06-01 00:00:00'),
11: Timestamp('2016-06-01 00:00:00'),
12: Timestamp('2016-07-01 00:00:00'),
13: Timestamp('2016-07-01 00:00:00'),
14: Timestamp('2016-08-01 00:00:00'),
15: Timestamp('2016-08-01 00:00:00'),
16: Timestamp('2016-09-01 00:00:00'),
17: Timestamp('2016-09-01 00:00:00'),
18: Timestamp('2016-10-01 00:00:00'),
19: Timestamp('2016-10-01 00:00:00'),
20: Timestamp('2016-11-01 00:00:00'),
21: Timestamp('2016-11-01 00:00:00'),
22: Timestamp('2016-12-01 00:00:00'),
23: Timestamp('2016-12-01 00:00:00'),
24: Timestamp('2017-01-01 00:00:00'),
25: Timestamp('2017-01-01 00:00:00')},
'src': {0: 'YYZ',
1: 'DFW',
2: 'YYZ',
3: 'DFW',
4: 'YYZ',
5: 'DFW',
6: 'YYZ',
7: 'DFW',
8: 'YYZ',
9: 'DFW',
10: 'YYZ',
11: 'DFW',
12: 'YYZ',
13: 'DFW',
14: 'YYZ',
15: 'DFW',
16: 'YYZ',
17: 'DFW',
18: 'YYZ',
19: 'DFW',
20: 'YYZ',
21: 'DFW',
22: 'YYZ',
23: 'DFW',
24: 'YYZ',
25: 'DFW'}}
I want the percentile per group of src and dest pairs. So there should only be 1 percentile value for each pair. I only want to perform the percentile given b where date = 2017-01-01 for each src and dest pair over the entire column a for each pair. Make sense?
I can do this manually for example for a specific pair i.e. src=YYZ and dest=SFT:
from scipy import stats
import datetime as dt
import pandas as pd
p0 = dt.datetime(2017,1,1)
# lets slice df for src=YYZ and dest = SFO
x = df[(df.src =='YYZ') &
(df.dest =='SFO') &
(df.dt ==p0)].b.values[0]
# given B, what percentile does it fall in for the entire column A for YYZ, SFO
stats.percentileofscore(df['a'],x)
61.53846153846154
In the above case, I did this manually for pairs YYZ and SFO. However, I have thousands of pairs in my df.
How do I vectorize this using pandas features rather than looping through every pair?
There must be a way to use groupby and use apply over a function?
My desired df should look something like:
src dest percentile
0 YYZ SFO 61.54
1 DFW PDX 23.07
2 XXX YYY blahblah1
3 AAA BBB blahblah2
...
UPDATE:
I implemented the following:
def b_percentile_a(df,x,y,b):
z = df[(df['src'] == x ) & (df['dest'] == y)].a
r = stats.percentileofscore(z,b)
return r
b_vector_df = df[df.dt == p0]
b_vector_df['p0_a_percentile_b'] = \
b_vector_df.apply(lambda x: b_percentile_a(df,x.src,x.dest,x.b), axis=1)
It takes 5.16 seconds for 100 pairs. I have 55,000 pairs. So this will take ~50 minutes. I need to run this 36 times so its going to take several days of run time.
There must be a faster approach?
回答1:
Obtained a incredible saving of time!
Output:
Size of a_list: 49998 Randomized unique values
percentile_1 (Your given df - scipy)
computed percentile 104 times - 104 records in 0:00:07.777022
percentile_9 (class PercentileOfScore(rank_searchsorted_list) using given df)
computed percentile 104 times - 104 records in 0:00:00.000609_ dt src dest a b pct scipy _
0: 2016-01-01 YYZ SFO 54812 279.28 74.81299251970079 74.8129925197
1: 2016-01-01 DFW PDX 111.35 -65.5 24.66698667946718 24.6669866795
2: 2016-02-01 YYZ SFO 64.84 342.35 76.4810592423697 76.4810592424
3: 2016-02-01 DFW PDX 63.81 61.64 63.84655386215449 63.8465538622
...
24: 2017-01-01 YYZ SFO 97.04 338.28 76.3570542821712 76.3570542822
25: 2017-01-01 DFW PDX 133.94 -129.69 21.4668586743469 21.4668586743
Looking at the implementation of scipy.percentileofscore i found that the whole list( a )
are - copied, inserted, sorted, searched - on every call of percentileofscore.
I implemented my own class PercentileOfScore
import numpy as np
class PercentileOfScore(object):
def __init__(self, aList):
self.a = np.array( aList )
self.a.sort()
self.n = float(len(self.a))
self.pct = self.__rank_searchsorted_list
# end def __init__
def __rank_searchsorted_list(self, score_list):
adx = np.searchsorted(self.a, score_list, side='right')
pct = []
for idx in adx:
# Python 2.x needs explicit type casting float(int)
pct.append( (float(idx) / self.n) * 100.0 )
return pct
# end def _rank_searchsorted_list
# end class PercentileOfScore
I don't think that def percentile_7 will fit your needs. dt will not considered.
PctOS = None
def percentile_7(df_flat):
global PctOS
result = {}
for k in df_flat.pair_dict.keys():
# df_flat.pair_dict = { 'src.dst': [b,b,...bn] }
result[k] = PctOS.pct( df_flat.pair_dict[k] )
return result
# end def percentile_7
In your manual sample you use the whole df.a. In this sample its dt_flat.a_list, but i'm not sure if this is what you want?
from PercentileData import DF_flat
def main():
# DF_flat.data = {'dt.src.dest':[a,b]}
df_flat = DF_flat()
# Instantiate Global PctOS
global PctOS
# df_flat.a_list = [a,a,...an]
PctOS = PercentileOfScore(df_flat.a_list)
result = percentile_7(df_flat)
# result = dict{'src.dst':[pct,pct...pctn]}
Tested with Python:3.4.2 and 2.7.9 - numpy: 1.8.2
回答2:
Assuming you have a list of pairs, say pairs = [[a,b], [c,d], ...] and df is defined,
r = stats.percentileofscore(z,b)
return r
for pair in pairs:
# get the corresponding rows for each pair
bvalues = df.loc[(df['src']==pair[0])&(df['dest']==pair[1])][['a', 'b']]
# apply the percentileofscore map
b_vector_df['p0_a_percentile_b'] = bvalues.b.apply(lambda x: stats.percentileofscore(bvalues.a, x))
I am not entirely sure what the goal is. My understanding is you read a b value for each src, dest pair and look for the corresponding a value and then calculate the percentile of that a value. Let me know if this helps :)
EDIT: assuming you are only working with the five columns date, src, dest, a, and b, you can consider working with a copy of the data frame that only contains those 5 columns. It reduces the amount of work required by each extraction step. I feel like it is more efficient to work with only the amount of data you need.
Selecting rows from a Dataframe based on values in multiple columns in pandas is a discussion that may be relevant for you.
回答3:
You can groupby multiple columns at once.
# takes the b value at a specified point
# and returns its percentile of the full a array
def b_pct(df, p0):
bval = df.b[df.dt==p0]
assert bval.size == 1, 'can have only one entry per timestamp'
bval = bval.values[0]
# compute the percentile
return (df.a < bval).sum() / len(df.a)
# splits the full dataframe up into groups by (src, dest) trajectory and
# returns a dataframe of the form src, dest, percentile
def trajectory_b_percentile(df, p0):
percentile_df = pd.DataFrame([pd.Series([s, d, b_pct(g, p0)],
index=['src', 'dest', 'percentile'])
for ((s, d), g) in df.groupby(('src', 'dest'))])
return percentile_df
For comparison, your code above spits out
dt src dest a b p0_a_percentile_b
24 2017-01-01 YYZ SFO 97.04 338.28 23.076923
25 2017-01-01 DFW PDX 133.94 -129.69 46.153846
whereas `trajectory_b_percentile' returns
src dest percentile
0 DFW PDX 46.1538
1 YYZ SFO 23.0769
I didn't see any speedup with 25 entries, but it should be noticeable with more.
回答4:
It seems another considerable speedup is obtained by converting everything to numpy arrays and constructing the percentiles also as a numpy array:
# Get airport strings as indices
_, ir = np.unique(df['src'].values, return_inverse=True)
_, ic = np.unique(df['dest'].values, return_inverse=True)
# Get a and b columns
a = df['a'].values
b = df['b'].values
# Compute percentile scores in a numpy array
prc = np.zeros(a.shape)
for i in range(0, a.shape[0]):
prc[i] = stats.percentileofscore(a[np.logical_and(ir==ir[i], ic==ic[i])], b[i])
On a dataframe with 24000 entries (see construction below), running %%timeit gives
1 loop, best of 3: 2.17 s per loop
However, the original version
df['p0_a_percentile_b'] = \
df.apply(lambda x: b_percentile_a(df,x.src,x.dest,x.b), axis=1)
yields
1 loop, best of 3: 1min 2s per loop
which is much slower. I also checked that both snippets produce the same output by running np.all(prc == df.p0_a_percentile_b.values), yielding True.
Appendix:
I constructed a dataframe to test this and here I share the process for reproducibility. I took 2000 pairs of airports using 100 unique airport names, then generated 12 dataframe rows per pair, and then generated random a and b columns.
import pandas as pd
import numpy as np
import scipy.stats as stats
import numpy.matlib as mat
# Construct dataframe
T=12
N_airports = 100
N_entries = 2000
airports = np.arange(0, N_airports).astype('string')
src = mat.repmat(airports[np.random.randint(N_airports, size=(N_entries, ))], 1, T)
dest = mat.repmat(airports[np.random.randint(N_airports, size=(N_entries, ))], 1, T)
a = np.random.uniform(size=N_entries*T)
b = np.random.uniform(size=N_entries*T)
df = pd.DataFrame(np.vstack((src, dest, a, b)).T, columns=['src', 'dest', 'a', 'b'])
回答5:
Please verify and comment if this represent your Data Model!
- 6 ^ 6 Pairs [AAA-ZZZ] = 46,656 are used. Typically each PAIR has 12 RECORDS
- This is RECORD( 0 ) of PAIR( DFW PDX )
dt src dest a b 0: 2016-01-01 DFW PDX 111.35 -65.5 - This is SET( DFW PDX ) = 13 RECORDS of PAIR( DFW PDX )
dt src dest a b 0: 2016-01-01 DFW PDX 111.35 -65.5 1: 2016-02-01 DFW PDX 63.81 61.64 2: 2016-03-01 DFW PDX -207.49 115.89 3: 2016-04-01 DFW PDX -56.43 2.35 4: 2016-05-01 DFW PDX -296.62 62.12 5: 2016-06-01 DFW PDX -376.26 -142.78 6: 2016-07-01 DFW PDX -821.73 -214.36 7: 2016-08-01 DFW PDX -1410.09 -207.62 8: 2016-09-01 DFW PDX -326.26 -132.53 9: 2016-10-01 DFW PDX -24.44 13.47 10:2016-11-01 DFW PDX -5.24 -144.6 11:2016-12-01 DFW PDX 195.21 -50.2 12:2017-01-01 DFW PDX 133.94 -129.69 - Example: Calculate Percentile of RECORD( 0 )
dt src dest a b 0: 2016-01-01 DFW PDX 111.35 -65.5Pseudocode: stats.percentileofscore( SET( DFW PDX )[a0...a12], -65.5) = 46.15
Example: Calculate Percentile of SET( DFW PDX )
Pseudocode
for record in SET( DFW PDX ):
stats.percentileofscore( SET( DFW PDX )[a0...a12], record.b)
Output: pct0...pct12Using rank_searchsorted_list does not require 'for record in' :
rank_searchsorted_list( SET( DFW PDX )[a0...a12], SET( DFW PDX )[b0...b12] )
Output: [pct0...pct12]This is SET( DFW PDX ) vectorized
OBJECT = {'DFW PDX':[ ['2016-01-01', '2016-02-01', '2016-03-01', '2016-04-01', '2016-05-01', '2016-06-01', '2016-07-01', '2016-08-01', '2016-09-01', '2016-10-01', '2016-11-01', '2016-12-01', '2017-01-01'] [111.35, 63.81, -207.49, -56.43, -296.62, -376.26, -821.73, -1410.09, -326.26, -24.44, -5.24, 195.21, 133.94] [-65.5, 61.64, 115.89, 2.35, 62.12, -142.78, -214.36, -207.62, -132.53, 13.47, -144.6, -50.2, -129.69] [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] ]}Example: Calculate Percentile of OBJECT( DFW PDX )
Using stats.percentileofscore:a = 1; b = 2 for b_value in OBJECT['DFW PDX'][b]: stats.percentileofscore( OBJECT['DFW PDX'][a], b_value) Output: pct0...pct12Using rank_searchsorted_list does not require 'for b_value in':
a = 1; b = 2; pct = 3 vector = OBJECT['DFW PDX'] vector[pct] = rank_searchsorted_list( vector[a], vector[b] )Output:
dt src dest a b pct scipy 0: 2016-01-01 DFW PDX 111.35 -65.5 46.15 46.15 1: 2016-02-01 DFW PDX 63.81 61.64 69.23 69.23 2: 2016-03-01 DFW PDX -207.49 115.89 84.61 84.61 3: 2016-04-01 DFW PDX -56.43 2.35 69.23 69.23 4: 2016-05-01 DFW PDX -296.62 62.12 69.23 69.23 5: 2016-06-01 DFW PDX -376.26 -142.78 46.15 46.15 6: 2016-07-01 DFW PDX -821.73 -214.36 38.46 38.46 7: 2016-08-01 DFW PDX -1410.09 -207.62 38.46 38.46 8: 2016-09-01 DFW PDX -326.26 -132.53 46.15 46.15 9: 2016-10-01 DFW PDX -24.44 13.47 69.23 69.23 10:2016-11-01 DFW PDX -5.24 -144.6 46.15 46.15 11:2016-12-01 DFW PDX 195.21 -50.2 53.84 53.84 12:2017-01-01 DFW PDX 133.94 -129.69 46.15 46.15
Please verify and confirm the calculated percentile!
来源：https://stackoverflow.com/questions/42076126/vectorize-percentile-value-of-column-b-of-column-a-for-groups