连续检查两个 True(布尔值 1)值的最快和/或最 pythonic 方法是什么?

Question

我想出了这个解决方案，但我想知道是否有一些内置函数可以更好，更快或以更python的方式执行此操作

import numpy as np
import pandas as pd
n = 1 #number of trials
p = 0.5 #prob of succes
k = 50 #amount of reptitions
s = pd.Series(np.random.binomial(n, p, k)).to_frame()
s.columns = ['data']
s['shifted'] = s['data'].shift(1)
s['lagged'] = s['data'].shift(-1)
s['two_ones_in_a_row'] = (s['data'] & s['lagged']) | (s['data'] & s['shifted'])

Answer 1

在我看来，解决方案很好，也不需要新列，您可以按Series进行比较：

s = pd.Series(np.random.binomial(n, p, k)).to_frame()
s.columns = ['data']
s1= s['data'].shift(1)
s2 = s['data'].shift(-1)
s['two_ones_in_a_row'] = (s['data'] & s2) | (s['data'] & s1)

如果性能很重要，请使用 numpy：

a = s['data'].values.astype(bool)
s['two_ones_in_a_row1'] = (a & np.append(a[1:], False)) | (a & np.append(False, a[:-1]))

---

n = 1 #number of trials
p = 0.5 #prob of succes
k = 50000 #amount of reptitions
s = pd.Series(np.random.binomial(n, p, k)).to_frame()
s.columns = ['data']

In [153]: %%timeit 
...: s1= s['data'].shift(1)
...: s2 = s['data'].shift(-1)
...: s['two_ones_in_a_row'] = (s['data'] & s2) | (s['data'] & s1)
...: 
21 ms ± 581 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [154]: %%timeit
...: a = s['data'].values.astype(bool)
...: s['two_ones_in_a_row1'] = (a & np.append(a[1:], False)) | (a & np.append(False, a[:-1]))
...: 
213 µs ± 2.92 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)