如何计算 Pandas DataFrame 中行之间的差异？

Question

在一个数据框中，我有 4 个变量，分别是机器人的 X、Y、Z 和 W 方向。每条线代表这四个值的测量值。

x = [-0.75853, -0.75853, -0.75853, -0.75852]
y = [-0.63435, -0.63434, -0.63435, -0.63436]
z = [-0.10488, -0.10490, -0.10492, -0.10495]
w = [-0.10597, -0.10597, -0.10597, -0.10596]

df = pd.DataFrame([x, y, z, w], columns=['x', 'y', 'z', 'w'])

我在下面写了函数 returns 两个四元数之间的三个区别：

from pyquaternion import Quaternion

def quaternion_distances(w1, x1, y1, z1, w2, x2, y2, z2):
    """ Create two Quaternions objects and calculate 3 distances between them """
    q1 = Quaternion(w1, x1, y1, z1)
    q2 = Quaternion(w2, x2, y2, z2)

    dist_by_signal  = Quaternion.absolute_distance(q1, q2)
    dist_geodesic   = Quaternion.distance(q1, q2)
    dist_sim_geodec = Quaternion.sym_distance(q1, q2)

    return dist_by_signal, dist_geodesic, dist_sim_geodec

这个差是根据第二行的值乘以第一行的值计算出来的。因此，我无法使用 Pandas 应用函数。

我已经向数据框添加了三列，因此我收到了函数返回的每个值：

df['dist_by_signal']  = 0
df['dist_geodesic']   = 0
df['dist_sim_geodec'] = 0

问题是：如何将上述函数应用于每一行并将结果包含在这些新列中？你能给我一个建议吗？

Answer 1

可以使用。

Quaternions=df.apply(lambda x: Quaternion(x), axis=1)
df['dist_by_signal']  = 0
df['dist_geodesic']   = 0
df['dist_sim_geodec'] = 0
df.reset_index(drop=True)
for i in df.index:
    q1=Quaternions[i]
    if i+1<len(df.index):
        q2=Quaternions[i+1]
    df.loc[i,['dist_by_signal','dist_geodesic','dist_sim_geodec']]=[Quaternion.absolute_distance(q1, q2), Quaternion.distance(q1, q2),Quaternion.sym_distance(q1, q2)]
print(df)

         x        y        z        w  dist_by_signal  dist_geodesic  \
0 -0.75853 -0.75853 -0.75853 -0.75852        0.248355       0.178778   
1 -0.63435 -0.63434 -0.63435 -0.63436        1.058875       1.799474   
2 -0.10488 -0.10490 -0.10492 -0.10495        0.002111       0.010010   
3 -0.10597 -0.10597 -0.10597 -0.10596        0.000000       0.000000   

   dist_sim_geodec  
0         0.178778  
1         1.799474  
2         0.010010  
3         0.000000

Answer 2

考虑 shift to create adjacent columns, w2, x2, y2, z2, of next row values then run rowwise apply 这确实需要 axis='columns'（不是 index）：

df[[col+'2' for col in list('wxyz')]] = df[['x', 'y', 'z', 'w']].shift(-1)

def quaternion_distances(row):

    """ Create two Quaternions objects and calculate 3 distances between them """ 
    q1 = Quaternion(row['w'], row['x'], row['y'], row['z'])
    q2 = Quaternion(row['w2'], row['x2'], row['y2'], row['z2'])

    row['dist_by_signal']  = Quaternion.absolute_distance(q1, q2)
    row['dist_geodesic']   = Quaternion.distance(q1, q2)
    row['dist_sim_geodec'] = Quaternion.sym_distance(q1, q2)

    return row


df = df.apply(quaternion_distances, axis='columns')

print(df)

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