直接翻譯你的代碼到一個列表的理解是:
positions = [obSet['results'][i]['location'] for i in range(obSet['total_results'])]
的obSet['total_results']
是言之有物,但沒有必要,你可以在obSet['results']
只是循環而直接使用每個結果詞典:
positions = [res['location'] for res in obSet['results']]
現在你有一個字符串列表但是,因爲每個'location'
仍然是long,lat
之前打印的格式化字符串。
拆分該字符串,並將結果轉換成浮動的序列:
positions = [map(float, res['location'].split(',')) for res in obSet['results']]
現在你有名單的浮點值的列表:
>>> [map(float, res['location'].split(',')) for res in obSet['results']]
[[9.5142456535, -83.8011438905], [10.2335478381, -84.8517773638], [10.3358965682, -84.9964271008], [10.3744851815, -84.9871494128], [10.2468720343, -84.9298072822], [10.3456659939, -84.9451804822], [10.3611732346, -84.9450302597], [10.3174360636, -84.8798676791], [10.325110706, -84.939710318], [9.4098152454, -83.9255607577], [9.4907141714, -83.9240819199], [9.562637289, -83.8170178428], [9.4373885911, -83.8312881263], [9.4766746409, -83.8120952573], [10.2651190176, -84.6360466565], [9.6572995298, -83.8322965118], [9.6997991784, -83.9076919066], [9.6811177044, -83.8487647156], [9.7416717045, -83.929327673], [9.4885099275, -83.9583968683], [10.1233252667, -84.5751029683], [9.4411815757, -83.824401543], [9.4202687169, -83.9550344212], [9.4620656621, -83.665183105], [9.5861809119, -83.8358881552], [9.4508914243, -83.9054016165], [9.4798058284, -83.9362558497], [9.5970449879, -83.8969131893], [9.5855562829, -83.8354434596], [10.2366179555, -84.854847472], [9.718459702, -83.8910277016], [9.4424384874, -83.8880459793], [9.5535916157, -83.9578166199], [10.4124554163, -84.9796942349], [10.0476688795, -84.298227929], [10.2129436252, -84.8384097435], [10.2052632717, -84.6053701877], [10.3835784147, -84.8677930134], [9.6079669672, -83.9084281155], [10.3583643315, -84.8069762134], [10.3975986735, -84.9196996767], [10.2060835381, -84.9698814407], [10.3322929317, -84.8805587129], [9.4756504472, -83.963818143], [10.3997876964, -84.9127311339], [10.1777433853, -84.0673088686], [10.3346128571, -84.9306278215], [9.5193346195, -83.9404786293], [9.421538224, -83.7689452093], [9.430427837, -83.9532672942], [10.3243212895, -84.9653175843], [10.021698503, -83.885674888]]
如果必須有元組而不是名單,增加一個tuple()
電話:
positions = [tuple(map(float, res['location'].split(',')))
for res in obSet['results']]
後者還確保表達式在Python 3中工作(其中map()
返回迭代器,而不是列表);否則你不得不使用嵌套列表理解:
# produce a list of lists in Python 3
positions = [[float(p) for p in res['location'].split(',')] for res in obSet['results']]
該json中的位置是*字符串*。 –
另外,由於你是Python新手,是否有任何特別的理由讓你使用Python 2,現在Python 3已經出現了** 8年**? –
呃,優點!我可能最好把它弄清楚。 –