已知15幅影像的内外方位元素，341个地面点，每个点至少6幅影像上的像点坐标。根据摄站中心、地面点及对应像点间的共线方程式，采用间接平差方法编程求解这些地面点的物方坐标，要求使用Python。

以下是一个Python实现的参考代码，采用最小二乘法进行间接平差计算：

import numpy as np

# 影像数
n = 15

# 地面点数
m = 341

# 初始化矩阵
A = np.zeros((m * 2, n + 3))
L = np.zeros((m * 2, 1))

# 像点坐标数据，每行6个像点坐标
X = np.array([
    [x1, y1, x2, y2, x3, y3, x4, y4, x5, y5, x6, y6, x7, y7],
    [x1, y1, x2, y2, x3, y3, x4, y4, x5, y5, x6, y6, x7, y7],
    ...
    [x1, y1, x2, y2, x3, y3, x4, y4, x5, y5, x6, y6, x7, y7]
])

# 内外方位元素数据，每行13个元素
P = np.array([
    [w1, k1, f1, x01, y01, z01, om1, phi1, kap1, x1, y1, z1, s1, t1],
    [w2, k2, f2, x02, y02, z02, om2, phi2, kap2, x2, y2, z2, s2, t2],
    ...
    [w15, k15, f15, x015, y015, z015, om15, phi15, kap15, x15, y15, z15, s15, t15]
])

# 地面点坐标数据，每行3个坐标
G = np.array([
    [X1, Y1, Z1],
    [X2, Y2, Z2],
    ...
    [X341, Y341, Z341]
])

# 首先计算出每个像点的像空间坐标
for i in range(m):
    for j in range(n):
        w, k, f, x0, y0, z0, om, phi, kap, x, y, z, s, t = P[j]
        Xw = np.array([[x], [y], [z]])
        R = np.array([
            [-np.sin(phi) * np.sin(kap) * np.sin(om) + np.cos(phi) * np.cos(om), -np.sin(phi) * np.sin(kap) * np.cos(om) - np.cos(phi) * np.sin(om), -np.sin(phi) * np.cos(kap)],
            [np.cos(phi) * np.sin(kap) * np.sin(om) + np.sin(phi) * np.cos(om), np.cos(phi) * np.sin(kap) * np.cos(om) - np.sin(phi) * np.sin(om), np.cos(phi) * np.cos(kap)],
            [np.cos(kap) * np.sin(om), -np.cos(kap) * np.cos(om), np.sin(kap)]
        ])
        Xc = R.dot(Xw - np.array([[x0], [y0], [z0]]))
        fX = f * Xc[0, 0] / Xc[2, 0] + s
        fY = f * Xc[1, 0] / Xc[2, 0] + t
        A[i * 2, j] = f / Xc[2, 0]
        A[i * 2, n] = -fX / Xc[2, 0]
        A[i * 2, n + 1] = -fY / Xc[2, 0]
        A[i * 2, n + 2] = fX * Xc[1, 0] / Xc[2, 0] - fY * Xc[0, 0] / Xc[2, 0]
        A[i * 2 + 1, j] = 0
        A[i * 2 + 1, n] = f / Xc[2, 0]
        A[i * 2 + 1, n + 1] = -fY / Xc[2, 0]
        A[i * 2 + 1, n + 2] = fY * Xc[1, 0] / Xc[2, 0]

# 计算L矩阵
for i in range(m):
    L[i * 2, 0] = X[i, 0]
    L[i * 2 + 1, 0] = X[i, 1]

# 进行最小二乘计算
AT = np.transpose(A)
ATA = AT.dot(A)
ATL = AT.dot(L)
N = np.linalg.inv(ATA)
X = N.dot(ATL)

# 输出结果
print(X)

其中，X矩阵即为求解出的地面点的物方坐标。注意，代码中的变量名和数据格式需要根据实际情况进行修改。