Question

我有一个字典

{country: [coordinate_set1,
           coordinate_set2,
           coordinate_set3]}

其中每个coordinate_set当前是一串长的坐标对。

如何访问每个coordinate_set并将其拆分为每对x_coord;y_coord的列表

{country: [[set1_pair1][set1_pair2][set1_pair...],
           [set2_pair1][set2_pair2][set2_pair...],
           [set3_pair1][set3_pair2][set3_pair...]}

然后最后，在这些列表中，创建每对x_coord, y_coord

的元组

{country: [[x1_1,y1_1][x1_2,y1_2][x1_...,y1_],
           [x2_1,y2_1][x2_2,y2_2][x2_...,y2_],
           [x3_1,y3_1][x3_2,y3_2][x3_...,y3_]}

由于这些值是以空格分隔和以分号分隔的，因此在每个字符串元素上使用.split(" ")和.split(";")应该相当简单，但我无法弄清楚如何索引和保存拆分

我觉得它应该像

那样

for country, coordinate in dict.items():
    for coordinate_set in coordinate:
        split_set = coordinate_set.split(sep = " ")
        for xy_set in split_set:
            xy_tuple = tuple(xy_set.split(sep = ";")

但我无法真正了解如何保存这些内容。

示例数据：编辑：已更新样本数据

Antarctica
-80.0401787251;-59.5720946926 -80.5496566711;-59.865849372
-79.4970594217;-159.20818356 -79.634208673;-161.127601285
-78.0470696006;-45.1547576564 -78.4781027223;-43.9208278062

Answer 1

这应该这样做，列表和字典理解的组合：

{country: [tuple(xy_set.split(';')) for cordinate_set in coordinate for xy_set in in cordinate_set.split()] for country, coordinate in dict.items()}

101方式：

result = {}
for for country, coordinate in dict.items():
    for cordinate_set in coordinate:
        for xy_set in in cordinate_set.split():
            e = tuple(xy_set.split(';'))
            if country in result:
                result[country].append(e)
            else:
                result[country] = [e]

Answer 2

Python 101方式

 dict1={'Antarctica':
    ['''-80.0401787251;-59.5720946926 -80.5496566711;-59.865849372
    -79.4970594217;-159.20818356 -79.634208673;-161.127601285
    -78.0470696006;-45.1547576564 -78.4781027223;-43.9208278062''']}
    for country, coordinate in dict1.items():
        cs=[]
        for coordinate_set in coordinate:
            split_set = coordinate_set.split(sep = " ")
            ss=[]
            for xy_set in split_set:
                xy_tuple = tuple(xy_set.split(sep = ";"))
                ss+=[xy_tuple]
            cs+=ss
        dict1[country]=cs

    print(dict1)

Answer 3

不使用理解和较少的代码行进行微小修改

data={'Antarctica':
    ["""-80.0401787251;-59.5720946926 -80.5496566711;-59.865849372 
    -79.4970594217;-159.20818356 -79.634208673;-161.127601285 
    -78.0470696006;-45.1547576564 -78.4781027223;-43.9208278062"""],'Ameriica':
    ["""-80.0401787251;-59.5720946926 -80.5496566711;-59.865849372 
    -79.4970594217;-159.20818356 -79.634208673;-161.127601285 
    -78.0470696006;-45.1547576564 -78.4781027223;-43.9208278062"""]}



for key,value in data.items():
      #print value
      updatevalue=[]
      inlist=value[0].split()

      for i in inlist:
            updatevalue.append(tuple(i.split(';')))
      data.update({key:updatevalue})

print data

在字典中就地分割列表和子列表

3 个答案: