距离矩阵半正矢

发布于 2025-01-18 04:55:41 字数 743 浏览 1 评论 0原文

我正在研究一个看起来像这样的数据框：

             lat       lon
id_zone
0        40.0795  4.338600
1        45.9990  4.829600
2        45.2729  2.882000
3        45.7336  4.850478
4        45.6981  5.043200

我正在尝试制作一个 Haverisne 距离矩阵。基本上对于每个区域，我想计算它与数据框中所有其他区域之间的距离。所以对角线上应该只有0。这是我使用的半正矢函数，但我无法制作矩阵。

def haversine(x):
    x.lon, x.lat, x.lon2, x.lat2 = map(radians, [x.lon, x.lat, x.lon2, x.lat2])
    # formule de Haversine
    dlon = x.lon2 - x.lon
    dlat = x.lat2 - x.lat
    a = sin(dlat / 2) ** 2 + cos(x.lat) * cos(x.lat2) * sin(dlon / 2) ** 2
    c = 2 * atan2(sqrt(a), sqrt(1 - a))
    km = 6367 * c
    return km

原文

I am working on a data frame that looks like this :

             lat       lon
id_zone
0        40.0795  4.338600
1        45.9990  4.829600
2        45.2729  2.882000
3        45.7336  4.850478
4        45.6981  5.043200

I'm trying to make a Haverisne distance matrix. Basically for each zone, I would like to calculate the distance between it and all the others in the dataframe. So there should be only 0s on the diagonal. Here is the Haversine function that I use but I can't make my matrix.

def haversine(x):
    x.lon, x.lat, x.lon2, x.lat2 = map(radians, [x.lon, x.lat, x.lon2, x.lat2])
    # formule de Haversine
    dlon = x.lon2 - x.lon
    dlat = x.lat2 - x.lat
    a = sin(dlat / 2) ** 2 + cos(x.lat) * cos(x.lat2) * sin(dlon / 2) ** 2
    c = 2 * atan2(sqrt(a), sqrt(1 - a))
    km = 6367 * c
    return km

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浮萍、无处依 2025-01-25 04:55:41

您可以使用该答案的解决方案 pandas”> pandas-创建来自数据框架的差异矩阵a>

或在您的特定情况下，在此示例中拥有类似数据框的地方：

             lat       lon
id_zone
0        40.0795  4.338600
1        45.9990  4.829600
2        45.2729  2.882000
3        45.7336  4.850478
4        45.6981  5.043200

您的函数定义为：

def haversine(first, second):
    # convert decimal degrees to radians
    lat, lon, lat2, lon2 = map(np.radians, [first[0], first[1], second[0], second[1]])

    # haversine formula
    dlon = lon2 - lon
    dlat = lat2 - lat
    a = np.sin(dlat/2)**2 + np.cos(lat) * np.cos(lat2) * np.sin(dlon/2)**2
    c = 2 * np.arcsin(np.sqrt(a))
    r = 6371 # Radius of earth in kilometers. Use 3956 for miles
    return c * r

您通过lat和lon > lon第一个位置和第二位置。

然后，您可以使用numpy创建一个距离矩阵，然后用haversine函数的距离结果替换零：

# create a matrix for the distances between each pair of zones
distances = np.zeros((len(df), len(df)))
for i in range(len(df)):
    for j in range(len(df)):
        distances[i, j] = haversine(df.iloc[i], df.iloc[j])
pd.DataFrame(distances, index=df.index, columns=df.index)

您的输出应与此相似：

id_zone           0           1           2           3           4
id_zone
0          0.000000  659.422944  589.599339  630.083979  627.383858
1        659.422944    0.000000  171.597296   29.555376   37.325316
2        589.599339  171.597296    0.000000  161.731366  174.983855
3        630.083979   29.555376  161.731366    0.000000   15.474533
4        627.383858   37.325316  174.983855   15.474533    0.000000

You can use the solution to this answer Pandas - Creating Difference Matrix from Data Frame

Or in your specific case, where you have a DataFrame like this example:

             lat       lon
id_zone
0        40.0795  4.338600
1        45.9990  4.829600
2        45.2729  2.882000
3        45.7336  4.850478
4        45.6981  5.043200

And your function is defined as:

def haversine(first, second):
    # convert decimal degrees to radians
    lat, lon, lat2, lon2 = map(np.radians, [first[0], first[1], second[0], second[1]])

    # haversine formula
    dlon = lon2 - lon
    dlat = lat2 - lat
    a = np.sin(dlat/2)**2 + np.cos(lat) * np.cos(lat2) * np.sin(dlon/2)**2
    c = 2 * np.arcsin(np.sqrt(a))
    r = 6371 # Radius of earth in kilometers. Use 3956 for miles
    return c * r

Where you pass the lat and lon of the first location and the second location.

You can then create a distance matrix using Numpy and then replace the zeros with the distance results from the haversine function:

# create a matrix for the distances between each pair of zones
distances = np.zeros((len(df), len(df)))
for i in range(len(df)):
    for j in range(len(df)):
        distances[i, j] = haversine(df.iloc[i], df.iloc[j])
pd.DataFrame(distances, index=df.index, columns=df.index)

Your output should be similar to this:

id_zone           0           1           2           3           4
id_zone
0          0.000000  659.422944  589.599339  630.083979  627.383858
1        659.422944    0.000000  171.597296   29.555376   37.325316
2        589.599339  171.597296    0.000000  161.731366  174.983855
3        630.083979   29.555376  161.731366    0.000000   15.474533
4        627.383858   37.325316  174.983855   15.474533    0.000000

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