Pandas 找到从每个点到其余点的距离，循环

我是python新手。 我有一个csv文件，在两列中包含400对x和y。 我想循环数据，使其从一对（x_I，y_I）开始，并找到该对与其余399个点之间的距离。我希望对所有对（x_I，y_I）重复该过程，并将结果附加到列表

Dist_I

import pandas as pd
x_y_data = pd.read_csv("x_y_points400_labeled_csv.csv")

x = x_y_data.loc[:,'x']
y = x_y_data.loc[:,'y']


i=0
j=0
while (i<len(x)):
    Dist=np.sqrt((x[i]-x)**2 + (y[j]-y)**2)
    i = 1 + i 
    j = 1 + j
print(Dist)

这是我走了多远，但这不是我想要得到的。我的目标是在所附图片中获得类似的内容。
提前谢谢你的帮助

---

您可以使用

广播

（

arr[：，None]

）一次完成此计算。这将为您提供所需的重复计算。否则，

scipy.space.distance.pdist将为您提供计算的上三角

样本数据



和西皮

from scipy.spatial.distance import pdist

pdist(df[['X', 'Y']])
array([2.8532972 , 0.82759587, 1.95770875, 3.00078036, 1.16534282,
       3.27316125, 2.91598992, 1.17270443, 1.70814458, 2.78266933,
       3.1214628 , 1.74902298, 3.7184812 , 1.77945856, 2.09245472])

将其转换为上述数据帧

L = len(df)

arr = np.zeros((L, L))     
arr[np.triu_indices(L, 1)] = pdist(df[['X', 'Y']])
arr = arr + arr.T # Lower triangle b/c symmetric

pd.DataFrame(arr, index=df.index, columns=df.index)

          point0    point1    point2    point3    point4    point5
point0  0.000000  2.853297  0.827596  1.957709  3.000780  1.165343
point1  2.853297  0.000000  3.273161  2.915990  1.172704  1.708145
point2  0.827596  3.273161  0.000000  2.782669  3.121463  1.749023
point3  1.957709  2.915990  2.782669  0.000000  3.718481  1.779459
point4  3.000780  1.172704  3.121463  3.718481  0.000000  2.092455
point5  1.165343  1.708145  1.749023  1.779459  2.092455  0.000000

非常感谢你的回答。所以我接下来尝试得到小于2的值，我尝试添加
result1=result
x = df['X'].to_numpy()
y = df['Y'].to_numpy()

result = pd.DataFrame(np.sqrt((x[:, None] - x)**2 + (y[:, None] - y)**2),
                      index=df.index,
                      columns=df.index)

          point0    point1    point2    point3    point4    point5
point0  0.000000  2.853297  0.827596  1.957709  3.000780  1.165343
point1  2.853297  0.000000  3.273161  2.915990  1.172704  1.708145
point2  0.827596  3.273161  0.000000  2.782669  3.121463  1.749023
point3  1.957709  2.915990  2.782669  0.000000  3.718481  1.779459
point4  3.000780  1.172704  3.121463  3.718481  0.000000  2.092455
point5  1.165343  1.708145  1.749023  1.779459  2.092455  0.000000

from scipy.spatial.distance import pdist

pdist(df[['X', 'Y']])
array([2.8532972 , 0.82759587, 1.95770875, 3.00078036, 1.16534282,
       3.27316125, 2.91598992, 1.17270443, 1.70814458, 2.78266933,
       3.1214628 , 1.74902298, 3.7184812 , 1.77945856, 2.09245472])

L = len(df)

arr = np.zeros((L, L))     
arr[np.triu_indices(L, 1)] = pdist(df[['X', 'Y']])
arr = arr + arr.T # Lower triangle b/c symmetric

pd.DataFrame(arr, index=df.index, columns=df.index)

          point0    point1    point2    point3    point4    point5
point0  0.000000  2.853297  0.827596  1.957709  3.000780  1.165343
point1  2.853297  0.000000  3.273161  2.915990  1.172704  1.708145
point2  0.827596  3.273161  0.000000  2.782669  3.121463  1.749023
point3  1.957709  2.915990  2.782669  0.000000  3.718481  1.779459
point4  3.000780  1.172704  3.121463  3.718481  0.000000  2.092455
point5  1.165343  1.708145  1.749023  1.779459  2.092455  0.000000