numpy 如何在列表中找到相似的数字序列并返回索引偏移量

jdg4fx2g  于 11个月前  发布在  其他
关注(0)|答案(1)|浏览(104)

给定两个相似但不相同的列表,我如何找到一个列表与另一个列表匹配的偏移量。它们是一列重叠的黑色和白色图像的像素值,因此它们的值接近但不精确。我根据它们的值将数字分组为3个“bin”(0,1,2)。我使用Python。在下面的列表中,列表A在列表B之前的44个位置(像素)开始。索引44之后的数字是“接近”但不精确的。预期的输出将是44。

列表

A [1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 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, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 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, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]

B [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 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, 0, 0, 0]

字符串

初始编码

import matplotlib.pyplot as plt
import numpy as np

A = [1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 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, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 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, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
B = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 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, 0, 0, 0]

def plot_this(aa, bb, idx, rating ):
    xs = [x for x in range(len(aa))]
    fig, ax = plt.subplots(figsize=(8, 4), layout='constrained')
    plt.plot(xs, aa, label ="A")
    plt.plot(xs, bb, label ="B")
    title = str(idx) + " = " + str(rating)
    ax.set_title(title)  # Add a title to the axes.
    ax.legend() 
    plt.show()


x = np.linspace(0, 2 * np.pi, 200)
y = np.sin(x)

fig, ax = plt.subplots()
ax.plot(x, y)
plt.show()
A1 = A.copy()
B1 = B.copy()
score = 0
bad = 0
score_dict = {}
for i in range(len(A1)):
    for j in range(len(A1)):
        if A1[j] == B1[j]:
            score += 1
        else:
            bad += 1
    if bad >= score:
        score_dict[i] = 0
    else:
        rating = (score-bad)/len(A)
        score_dict[i] = rating
        if rating > 0.51:
            plot_this(A1, B1, i, rating)
    score = 0
    bad = 0
    A1.pop(0)
    B1.pop(-1)
for kk, vv in score_dict.items():
    print(f"index {kk}  Score {vv}")

验证码

import matplotlib.pyplot as plt
import numpy as np

A = [1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 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, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 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, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
B = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 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, 0, 0, 0]
def plot_this(aa, bb, idx, rating ):
    xs = [x for x in range(len(aa))]
    fig, ax = plt.subplots(figsize=(8, 4), layout='constrained')
    plt.plot(xs, aa, label ="A")
    plt.plot(xs, bb, label ="B")
    title = str(idx) + " = " + str(rating)
    ax.set_title(title)  # Add a title to the axes.
    ax.legend() 
    plt.show()

A1 = A.copy()
B1 = B.copy()
score_dict = {}
for i in range(int(2*len(A1)/3)): #Only go 2/3 way
    coeff = np.corrcoef(A1, B1)
    ranking = coeff[0,1]
    score_dict[i] = coeff[0,1]
    if ranking > 0.9:
            plot_this(A1, B1, i, ranking)
    A1.pop(0) #pop off both ends instead of using numpy.roll()
    B1.pop(-1) #Use numpy.roll() and go all the way around if you don't know which way to go.
for kk, vv in score_dict.items():
    print(f"index {kk}  Score {vv}")
high_score = max(score_dict, key=score_dict.get)
print(high_score)

最终图

nzrxty8p

nzrxty8p1#

用numpy可以吗?那么我建议您使用numpy.roll()对第一个数组进行移位,使用numpy.corrcoef()计算数组之间的相关性。如果相关系数接近1,则表明两个数组之间的一致性很好。

相关问题