如何仅获取字符串中包含的列表中的第一个元素？

发布于 2025-01-22 17:40:15 字数 324 浏览 1 评论 0原文

我有一个长字符串。该字符串包含一个列表，例如

'[{"ex1": 0, "ex2":1}, {"ex3": 2, "ex4":3}]'

我可以使用json5.5.loads，然后通过在列表上使用[0]获得第一个元素，但是json5。 LOADS需要长时间的时间来完成更长的字符串。有没有办法仅在不加载整个列表的情况下获得第一个元素？（在此示例中，它将是{“ ex1”：0，“ ex2”：1}。逗号分裂对我不起作用，因为列表中的字典中包含逗号。谢谢。

原文

I've got a long string. This string contains a list, like such example

'[{"ex1": 0, "ex2":1}, {"ex3": 2, "ex4":3}]'

I can use json5.loads and then get the first element by using [0] on the list, but json5.loads takes a long time for longer strings. Is there a way to get just the first element without loading the entire list? (in this example it would be {"ex1": 0, "ex2":1}. Splitting by commas doesn't work for me since there are commas contained in dictionaries in the list. Thanks.

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待天淡蓝洁白时 2025-01-29 17:40:15

您的字符串是否可以与ast.literal_eval（）一起使用？如果是这样，您可以做到，

obj = ast.literal_eval(s)
# obj[0] gives the first dict

如果不这样做，当开放式支架的数量等于近距离托架的数量时，您可以循环循环并产生任何子字符串。

def get_top_level_dict_str(s):
  open_br = 0
  close_br = 0
  open_index = 0
  for i, c in enumerate(s):
    if c == '{':
        if open_br == 0: open_index = i 
        open_br += 1
    elif c == '}':
        close_br += 1
        if open_br > 0 and open_br == close_br:
            yield s[open_index:i+1]
            open_br = close_br = 0

如果要将所得的子字符串解析为对象，则可以像已经这样做的那样使用json5，在较小的字符串上可能会更快，或使用ast.literal_eval（）

x = get_top_level_dict_str(s)
# next(x) gives the substring
# then use json5 or ast.literal_eval()

Does your string work with ast.literal_eval()? If it does, you could do

obj = ast.literal_eval(s)
# obj[0] gives the first dict

If not, you could loop through the string character-by-character and yield any substring when the number of open-brackets are equal to the number of close-brackets.

def get_top_level_dict_str(s):
  open_br = 0
  close_br = 0
  open_index = 0
  for i, c in enumerate(s):
    if c == '{':
        if open_br == 0: open_index = i 
        open_br += 1
    elif c == '}':
        close_br += 1
        if open_br > 0 and open_br == close_br:
            yield s[open_index:i+1]
            open_br = close_br = 0

If you want to parse the resulting substrings to objects, you could use json5 like you already do, which is probably faster on the smaller string, or use ast.literal_eval()

x = get_top_level_dict_str(s)
# next(x) gives the substring
# then use json5 or ast.literal_eval()

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天涯沦落人 2025-01-29 17:40:15

如果肯定会是这种格式，则可以搜索开始和结束括号。

mystr = '[{"ex1": 0, "ex2":1}, {"ex3": 2, "ex4":3}]'
first = mystr.index("{")
last = mystr.index("}")
extracted = mystr[first:last+1]
print(extracted)

这是打印的
'{“ ex1”：0，“ ex2”：1}'

对于更复杂的字符串：

mystr = '[{"ex1": {"ex1.33": -1, "ex1.66": -2}, "ex2":1}, {"ex3": 2, "ex4":3}]'
n_open = 0
n_close = 0
first = mystr.index("{")
for ii in range(len(mystr)):
    if mystr[ii] == "{":
        n_open += 1
    elif mystr[ii] == "}":
        n_close += 1
    if n_open > 0 and n_open == n_close:
        break
extracted = mystr[first:ii+1]

If it'll definitely be that format, you can just search for the beginning and ending brackets.

mystr = '[{"ex1": 0, "ex2":1}, {"ex3": 2, "ex4":3}]'
first = mystr.index("{")
last = mystr.index("}")
extracted = mystr[first:last+1]
print(extracted)

this prints
'{"ex1": 0, "ex2":1}'

For a more complicated string:

mystr = '[{"ex1": {"ex1.33": -1, "ex1.66": -2}, "ex2":1}, {"ex3": 2, "ex4":3}]'
n_open = 0
n_close = 0
first = mystr.index("{")
for ii in range(len(mystr)):
    if mystr[ii] == "{":
        n_open += 1
    elif mystr[ii] == "}":
        n_close += 1
    if n_open > 0 and n_open == n_close:
        break
extracted = mystr[first:ii+1]

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