[英]Python build one dictionary from a list of keys, and a list of lists of values
[英]How to build a python dictionary from a list of keys and a list values, quickly?
我必须列出
labels = ['normal.']
percentages = [0.9936]
我想从这两个列表中建立字典
d = {}
for k, v in enumerate(lables, percentages):
d[k] = v
但是我得到了错误:
TypeError: 'list' object cannot be interpreted as an integer
这有什么问题吗?
编辑
然后当我收到字典时,我想执行此操作
result = [str(k) + ": " + str(v) for k, v in previous_dict]
一种方法是压缩两个列表一起,并压缩的对象转换为一个字典。 之后,您可以在dict.items()
上迭代以创建列表
In [158]: labels = ['normal.']
...: percentages = [0.9936]
In [159]: previous_dict = dict(zip(labels,percentages))
In [159]: previous_dict
Out[159]: {'normal.': 0.9936}
In [24]: result = [str(k) + ": " + str(v) for k, v in previous_dict.items()]
In [25]: result
Out[25]: ['normal.: 0.9936']
还枚举为您提供类型(index, element)
的元组列表,您不能像这样传递两个迭代器,但可以再次zip
这两个迭代器并制作一个字典,代码如下。
对于Python 3.6+,我们也可以使用f字符串格式化字符串
In [167]: labels = ['normal.']
...: percentages = [0.9936]
In [169]: d = {}
...: for k, v in zip(labels, percentages):
...: d[k] = v
In [170]: d
Out[170]: {'normal.': 0.9936}
In [30]: result = [f'{k}:{v}' for k, v in previous_dict.items()]
In [31]: result
Out[31]: ['normal.:0.9936']
这是您尝试使用列表理解的答案。 要使第二步正常工作,您需要使用.items()
来访问键和值
labels = ['normal.']
percentages = [0.9936]
previous_dict = {k:v for k, v in zip(labels, percentages)}
# {'normal.': 0.9936}
result = [str(k) + ": " + str(v) for k, v in previous_dict.items()]
# ['normal.: 0.9936']
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