嗨我有如下列表,其中包含来自图像的元数据,如下所示:
['Component 1: Y component: Quantization table 0, Sampling factors 1 horiz/1 vert',
'Component 2: Cb component: Quantization table 1, Sampling factors 1 horiz/1 vert',
'Component 3: Cr component: Quantization table 1, Sampling factors 1 horiz/1 vert',
'Compression Type: Progressive, Huffman', 'Content-Length: 14312', 'Content-Type: image/jpeg’]
我想使用以下格式拆分列表“:”来制作字典:
{Component 1: {Y component: [Quantization table 0, Sampling factors 1 horiz/1 vert’],
Component 2: {Cb component: [Quantization table 1, Sampling factors 1 horiz/1 vert]},
Component 3: {Cr component: [Quantization table 1, Sampling factors 1 horiz/1 vert]},
Compression Type: [Progressive, Huffman],Content-Length: 14312,Content-Type: image/jpeg}
目前我已经编写了一些无效的代码。
def make_dict(seq):
res = {}
if seq[0] is not '':
for elt in seq:
k, v = elt.split(':')
try:
res[k].append(v)
except KeyError:
res[k] = [v]
print res
此代码不起作用。我也尝试了其他方法,但我无法获得格式。
答案 0 :(得分:3)
您可以使用collections.OrderedDict
:
>>> li=['Component 1: Y component: Quantization table 0, Sampling factors 1 horiz/1 vert', 'Component 2: Cb component: Quantization table 1, Sampling factors 1 horiz/1 vert', 'Component 3: Cr component: Quantization table 1, Sampling factors 1 horiz/1 vert', 'Compression Type: Progressive, Huffman', 'Content-Length: 14312', 'Content-Type: image/jpeg']
>>> d=OrderedDict((sub[0],{sub[1]:sub[2:]}) if sub[2:] else (sub[0],sub[1]) for sub in [item.split(':') for item in li])
>>> d
OrderedDict([('Component 1', {' Y component': [' Quantization table 0, Sampling factors 1 horiz/1 vert']}), ('Component 2', {' Cb component': [' Quantization table 1, Sampling factors 1 horiz/1 vert']}), ('Component 3', {' Cr component': [' Quantization table 1, Sampling factors 1 horiz/1 vert']}), ('Compression Type', ' Progressive, Huffman'), ('Content-Length', ' 14312'), ('Content-Type', ' image/jpeg')])
>>>
答案 1 :(得分:3)
l = ['Component 1: Y component: Quantization table 0, Sampling factors 1 horiz/1 vert',
'Component 2: Cb component: Quantization table 1, Sampling factors 1 horiz/1 vert',
'Component 3: Cr component: Quantization table 1, Sampling factors 1 horiz/1 vert',
'Compression Type: Progressive, Huffman', 'Content-Length: 14312', 'Content-Type: image/jpeg']
d = {}
for ele in l:
spl = ele.split(":", 2)
if len(spl) == 3:
k1, k2, v = spl
d[k1] = {k2: v.split(",")}
else:
k,v = spl
d[k] = v.split() if "," in v else v
输出:
{'Component 1': {' Y component': [' Quantization table 0',
' Sampling factors 1 horiz/1 vert']},
'Component 2': {' Cb component': [' Quantization table 1',
' Sampling factors 1 horiz/1 vert']},
'Component 3': {' Cr component': [' Quantization table 1',
' Sampling factors 1 horiz/1 vert']},
'Compression Type': [' Progressive', ' Huffman'],
'Content-Length': ' 14312',
'Content-Type': ' image/jpeg'}
要删除空格,您可以str.strip
关闭它:
d = {}
for ele in l:
spl = ele.split(":", 2)
if len(spl) == 3:
k1, k2, v = spl
d[k1] = {k2.strip(): list(map(str.strip,v.split(",")))}
else:
k,v = spl
d[k] = list(map(str.strip, v.split())) if "," in v else v.strip
输出:
{'Component 1': {'Y component': ['Quantization table 0',
'Sampling factors 1 horiz/1 vert']},
'Component 2': {'Cb component': ['Quantization table 1',
'Sampling factors 1 horiz/1 vert']},
'Component 3': {'Cr component': ['Quantization table 1',
'Sampling factors 1 horiz/1 vert']},
'Compression Type': ['Progressive', 'Huffman'],
'Content-Length': '14312',
'Content-Type': 'image/jpeg'}
这两者实际上都符合您的预期输出。
答案 2 :(得分:2)
如果要处理任何级别的字典嵌套,可以使用如下所示的递归算法。示例 -
def makedict(elem):
if ':' in elem:
k,v = map(str.strip, elem.split(':',1))
return {k:makedict(v)}
elif ',' in elem:
elems = list(map(str.strip, elem.split(','))) #Simply map(...) for Python 2.x
return elems
return elem
如果你想制作一本字典词典,你可以这样做 -
d = {}
for elem in s:
d.update(makedict(elem))
或者,如果你想要一个词典词典列表,请在列表推导中为列表中的每个元素调用上面的函数,例如 -
result = [makedict(elem) for elem in yourlist]
词典词典演示 -
>>> d = {}
>>> for elem in s:
... d.update(makedict(elem))
...
>>> d
{'Component 2': {'Cb component': ['Quantization table 1', 'Sampling fac
>>> import pprint
>>> pprint.pprint(d)
{'Component 1': {'Y component': ['Quantization table 0',
'Sampling factors 1 horiz/1 vert']},
'Component 2': {'Cb component': ['Quantization table 1',
'Sampling factors 1 horiz/1 vert']},
'Component 3': {'Cr component': ['Quantization table 1',
'Sampling factors 1 horiz/1 vert']},
'Compression Type': ['Progressive', 'Huffman'],
'Content-Length': '14312',
'Content-Type': 'image/jpeg'}
答案 3 :(得分:1)
您可以使用递归优雅地解决问题,并且可以使用分割限制(split
的第二个参数来限制拆分计数):
def make_dict(l):
d = dict()
for elem in l:
key, value = elem.split(':', 1)
if ':' in value:
d[key] = make_dict([value])
else:
d[key] = value
return d
并且测试似乎符合您的期望:
>>> l = ['Component 1: Y component: Quantization table 0, Sampling factors 1 horiz/1 vert',
'Component 2: Cb component: Quantization table 1, Sampling factors 1 horiz/1 vert',
'Component 3: Cr component: Quantization table 1, Sampling factors 1 horiz/1 vert',
'Compression Type: Progressive, Huffman', 'Content-Length: 14312', 'Content-Type: image/jpeg']
>>> make_dict(l)
{'Component 1': {' Y component': ' Quantization table 0, Sampling factors 1 horiz/1 vert'},
'Component 2': {' Cb component': ' Quantization table 1, Sampling factors 1 horiz/1 vert'},
'Component 3': {' Cr component': ' Quantization table 1, Sampling factors 1 horiz/1 vert'},
'Compression Type': ' Progressive, Huffman',
'Content-Length': ' 14312',
'Content-Type': ' image/jpeg'}