資料如下所示：
d = {'location_id': [1, 2, 3, 4, 5], 'x': [47.43715, 48.213889, 46.631111, 46.551111, 47.356628], 'y': [11.880689, 14.274444, 14.371, 13.665556, 11.705181]}
df = pd.DataFrame(data=d)
print(df)
location_id x y
0 1 47.43715 11.880689
1 2 48.213889 14.274444
2 3 46.631111 14.371
3 4 46.551111 13.665556
4 5 47.356628 11.705181
預期輸出：
{(47.43715, 11.880689): 1, (48.213889, 14.274444): 2, (46.631111, 14.371): 3, ...}
所以我可以簡單地訪問提供點坐標的 ID。
我試過的：
dict(zip(df['x'].astype('float'), df['y'].astype('float'), zip(df['location_id'])))
Error: ValueError: dictionary update sequence element #0 has length 3; 2 is required
or
dict(zip(tuple(df['x'].astype('float'), df['y'].astype('float')), zip(df['location_id'])))
TypeError: tuple expected at most 1 arguments, got 2
我已經用谷歌搜索了一段時間，但我不是很清楚。感謝您提供任何幫助。
uj5u.com熱心網友回復：
我認為這
result = dict(zip(zip(df['x'], df['y']), df['location_id']))
應該給你你想要的嗎？結果：
{(47.43715, 11.880689): 1,
(48.213889, 14.274444): 2,
(46.631111, 14.371): 3,
(46.551111, 13.665556): 4,
(47.356628, 11.705181): 5}
uj5u.com熱心網友回復：
我沒有使用資料框，這是你想要的嗎？
my_dict = {}
d = {'location_id': [1, 2, 3, 4, 5], 'x': [47.43715, 48.213889, 46.631111, 46.551111, 47.356628], 'y': [11.880689, 14.274444, 14.371, 13.665556, 11.705181]}
for i in range(len(d['location_id'])):
my_dict[ (d['x'][i] , d['y'][i]) ] = d['location_id'][i]
uj5u.com熱心網友回復：
您可以將列設定為索引，然后將列匯出x到字典ylocation_id
d = df.set_index(['x', 'y'])['location_id'].to_dict()
print(d)
{(47.43715, 11.880689): 1, (48.213889, 14.274444): 2, (46.631111, 14.371): 3, (46.551111, 13.665556): 4, (47.356628, 11.705181): 5}
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