JSON格式化通过将字典值附加到列表来实现

Question

我有一个JSON对象，它是这样的：

{ "produktNr:"1234",
  "artNr_01":"12",
  "artNr_02":"23",
  "artNr_03":"",
  "artNr_04":"14",
  "name_01":"abc",
  "name_02":"der",
  "test":"junk"
}

我想把它转换成这样的字典：

{ "produktNr:"1234", "artNr":["12","23","","14"], "name":["abc","der"], "test":"junk"}

这种转换是基于给定的序列，例如seq = ["artNr","name"]。因此，在字典的键中搜索序列的内容，并将值收集到一个列表中。你知道吗

我目前的尝试：

tempDict = {}
for key,value in fmData.iteritems():
    for seqval in seq:
        if seqval in key:
            if seqval in tempDict:
                tempDict[seqval].append(value)
            else:
                x = []
                x.append(value)
                tempDict[seqval]=x
        else:
            tempDict[key] = value

面临一些问题。你知道吗

• 值列表没有顺序，即"artNr":["","14","12","23"] 而不是[_01,_02,_03,_04]

• 无法从字典中弹出项，因为在循环中无法删除字典项，从而导致：

  {“produktNr:”1234“，”artNr“：[”12“，”23“，”14“]，”artNr\u 01“：”12“，”artNr\u 02“：”23“，”artNr\u 03“，”artNr\u 04“：”14“，”name“：[”abc“，”der“，”name\u 01“：”abc“，”name\u 02“：”der“，”test“：”junk“}

我很想知道如何处理这个问题，特别是如果有一个Python的方法来解决这个问题。你知道吗

Answer 1

您可以使用OrderedDict包中的collections：

from collections import OrderedDict
import re

input_dict = { "produktNr":"1234",
               "artNr_01":"12",
               "artNr_02":"23",
               "artNr_03":"",
               "artNr_04":"14",
               "name_01":"abc",
               "name_02":"der",
               "test":"junk" }

# split keys on the first '_'
m = re.compile('^([^_]*)_(.*)')

def _order_by( item ):
    # helper function for ordering the dict.
    # item is split on first '_' and, if it was successful 
    # the second part is returned otherwise item is returned
    # if key is something like artNr_42, return 42
    # if key is something like test, return test 
    k,s = item
    try:
        return m.search(k).group(2)
    except:
        return k

# create ordered dict using helper function
orderedDict = OrderedDict( sorted(input_dict.items(), key=_order_by))

aggregated_dict = {}
for k, v in orderedDict.iteritems():
    # split key
    match = m.search(k)

    if match:
        # key is splittable, i.e., key is something like artNr_42
        kk = match.group(1)
        if kk not in aggregated_dict:
            # create list and add value
            aggregated_dict[kk] = [v]
        else:
            # add value
            aggregated_dict[kk].append(v)
    else:
        # key is not splittable, i.e., key is something like produktNr
        aggregated_dict[k] = v

print(aggregated_dict)

从而得到所需的输出

{'produktNr': '1234', 'test': 'junk', 'name': ['abc', 'der'], 'artNr': ['12', '23', '', '14']}

Answer 2

您可以重新创建一个新的字典，该字典将使用列表中键中的'_'对键的值进行分组，而其他键和值则保持不变。这应该做到：

d = { "produktNr":"1234", "artNr_01":"12", "artNr_02":"23","artNr_03":"","artNr_04":"14","name_01":"abc","name_02":"der","test":"junk"}
new_d= {}
for k, v in d.items():
    k_new = k.split('_')[0]
    if '_' in k:
        if k_new not in new_d:
            new_d[k_new] = [v]
        else:
            new_d[k_new].append(v)
    else:
        new_d[k_new] = v
print(new_d)
# {'artNr': ['', '14', '23', '12'], 'test': 'junk', 'produktNr': '1234', 'name': ['der', 'abc']}

dict是unordered集合，因此值附加到列表的顺序将是不确定的。你知道吗

Answer 3

对代码稍作修改：

tempDict = {}
for key,value in fmData.iteritems():
    seqval_in_key = "no"
    for seqval in seq:
        if seqval in key:
            seqval_in_key = "yes"
    for seqval in seq:
        if seqval in key:
            if seqval in tempDict:
                tempDict[seqval].append(value)
            else:
                x = []
                x.append(value)
                tempDict[seqval]=x
        else:
            if (seqval_in_key == "no"):
                tempDict[key] = value
print tempDict

结果：

{'produktNr': '1234', 'test': 'junk', 'name': ['abc', 'der'], 'artNr': ['14', '23', '', '12']}