Reputation: 1354
I have a data source which I can call to get end of day time-series data for financial instruments.
As an example, say that I have the following financial instruments' data already coded and retrieved from the data source:
price_a
[24.74636733, 29.65460993, 28.09686357, 16.24366395, 27.26716605, 17.1444073, 18.76608861, 17.68487362, 19.5026825, 25.62365151, 12.92619601, 25.66759065, 24.40646289, 15.61753458, 13.82584258, 27.2508518, 12.22547517, 24.2317834, 13.33257932, 28.18551972, 19.11053867, 10.43027953, 21.18221807, 15.1889216, 27.65876136, 16.72982501, 14.0134465, 22.68824162, 19.14317233, 13.57868721]
price_b
[21.01623084, 27.6426434, 20.16877846, 27.41341083, 23.39068249, 20.65973567, 28.11032189, 21.85843902, 20.26838929, 28.52493215, 24.11865407, 28.30861237, 20.51648305, 21.75927511, 21.82957788, 25.4647031, 25.4647031, 25.4647031, 25.4647031, 25.4647031, 25.4647031, 25.4647031, 25.4647031, 21.5721344, 20.41526114, 24.24593747, 25.23109812, 26.11780617, 25.13995547, 25.2511254]
days
['2016-06-01', '2016-06-02', '2016-06-03', '2016-06-04', '2016-06-05', '2016-06-06', '2016-06-07', '2016-06-08', '2016-06-09', '2016-06-10', '2016-06-11', '2016-06-12', '2016-06-13', '2016-06-14', '2016-06-15', '2016-06-16', '2016-06-17', '2016-06-18', '2016-06-19', '2016-06-20', '2016-06-21', '2016-06-22', '2016-06-23', '2016-06-24', '2016-06-25', '2016-06-26', '2016-06-27', '2016-06-28', '2016-06-29', '2016-06-30']
price_b
has duplicate data. Let's say that I know duplicate data only occurs in sets of 8 or more (such as in the example), and any number of repetitions less than 8 are just due by chance. What's a way to detect the sets of duplicate in price_b, and then using the index of the duplicate set, remove the price_a
data indexed from that set?
Expected output:
price_a
[24.74636733, 29.65460993, 28.09686357, 16.24366395, 27.26716605, 17.1444073, 18.76608861, 17.68487362, 19.5026825, 25.62365151, 12.92619601, 25.66759065, 24.40646289, 15.61753458, 13.82584258, 27.2508518, 15.1889216, 27.65876136, 16.72982501, 14.0134465, 22.68824162, 19.14317233, 13.57868721]
price_b
[21.01623084, 27.6426434, 20.16877846, 27.41341083, 23.39068249, 20.65973567, 28.11032189, 21.85843902, 20.26838929, 28.52493215, 24.11865407, 28.30861237, 20.51648305, 21.75927511, 21.82957788, 25.4647031, 21.5721344, 20.41526114, 24.24593747, 25.23109812, 26.11780617, 25.13995547, 25.2511254]
days
['2016-06-01', '2016-06-02', '2016-06-03', '2016-06-04', '2016-06-05', '2016-06-06', '2016-06-07', '2016-06-08', '2016-06-09', '2016-06-10', '2016-06-11', '2016-06-12', '2016-06-13', '2016-06-14', '2016-06-15', '2016-06-16', '2016-06-24', '2016-06-25', '2016-06-26', '2016-06-27', '2016-06-28', '2016-06-29', '2016-06-30']
Upvotes: 4
Views: 400
Reputation: 294488
price_a = [24.74636733, 29.65460993, 28.09686357, 16.24366395, 27.26716605,
17.1444073, 18.76608861, 17.68487362, 19.5026825, 25.62365151,
12.92619601, 25.66759065, 24.40646289, 15.61753458, 13.82584258,
27.2508518, 12.22547517, 24.2317834, 13.33257932, 28.18551972,
19.11053867, 10.43027953, 21.18221807, 15.1889216, 27.65876136,
16.72982501, 14.0134465, 22.68824162, 19.14317233, 13.57868721]
price_b = [21.01623084, 27.6426434, 20.16877846, 27.41341083, 23.39068249,
20.65973567, 28.11032189, 21.85843902, 20.26838929, 28.52493215,
24.11865407, 28.30861237, 20.51648305, 21.75927511, 21.82957788,
25.4647031, 25.4647031, 25.4647031, 25.4647031, 25.4647031,
25.4647031, 25.4647031, 25.4647031, 21.5721344, 20.41526114,
24.24593747, 25.23109812, 26.11780617, 25.13995547, 25.2511254]
days = ['2016-06-01', '2016-06-02', '2016-06-03', '2016-06-04', '2016-06-05',
'2016-06-06', '2016-06-07', '2016-06-08', '2016-06-09', '2016-06-10',
'2016-06-11', '2016-06-12', '2016-06-13', '2016-06-14', '2016-06-15',
'2016-06-16', '2016-06-17', '2016-06-18', '2016-06-19', '2016-06-20',
'2016-06-21', '2016-06-22', '2016-06-23', '2016-06-24', '2016-06-25',
'2016-06-26', '2016-06-27', '2016-06-28', '2016-06-29', '2016-06-30']
df = pd.DataFrame(dict(
a=price_a,
b=price_b,
), pd.to_datetime(days))
partitions = (df.b.diff() != 0).cumsum()
vc = partitions.value_counts()
vc8 = vc[vc == 8].index
df[~partitions.isin(vc8)]
Upvotes: 2