如何通过将列表的索引视为键来使包含列表的字典扁平化? [英] How can I flatten dict containing list by considering its index as key?

问题描述

我想把字典弄平.词典中可能包含列表.因此，在扁平化字典中的列表时，应将列表索引作为其键.

I want to flatten the dict. Dictionary may contain lists. So while flattening lists inside dictionary, it should consider list index as its key.

我该怎么做?

我尝试过:

def flatten(d, parent_key='', sep='__'):
    items = []
    for k, v in d.items():
        new_key = parent_key + sep + k if parent_key else k
        if isinstance(v, collections.MutableMapping):
            items.extend(flatten(v, new_key, sep=sep).items())
        else:
            items.append((new_key, v))
    return dict(items)

这是平整的字典，但忽略了列表.

This is flattening dict but ignoring lists.

我还尝试添加 if isinstance(v，list):，但是我没有在项目.

I also tried to add if isinstance(v, list):, but I am not getting how to append / extend in items.

data = {
    "checksum": "c540fcd985bf88c87e48c2bfa1df5498",
    "data": {
        "sampleMetrics": {
            "name": "DNA Library QC Metrics",
            "passQualityControl": "true",
            "metrics": [{
                "name": "CONTAMINATION_SCORE",
                "value": 1302,
                "LSL": 0,
                "USL": 3106,
                "UOM": "NA"
            }]
        }
    }
}
print flatten(data)

我得到的输出:

{
    'checksum': 'c540fcd985bf88c87e48c2bfa1df5498',
    'data__sampleMetrics__metrics': [{
        'LSL': 0,
        'USL': 3106,
        'name': 'CONTAMINATION_SCORE',
        'value': 1302,
        'UOM': 'NA'
    },{ 'demo': 11}],
    'data__sampleMetrics__name': 'DNA Library QC Metrics',
    'data__sampleMetrics__passQualityControl': 'true'
}

除了列表元素之外，还有什么其他扁平化的内容.

Which is flattening other thing except list elements.

预期输出:它也应该使列表变平.(通过将列表索引作为键.)

Expected output : It should flatten list as well.(By considering list index as key.)

{
    'checksum': 'c540fcd985bf88c87e48c2bfa1df5498',
    'data__sampleMetrics__metrics__0__LSL': 0,
    'data__sampleMetrics__metrics__0__USL': 3106,
    'data__sampleMetrics__metrics__0__name': 'CONTAMINATION_SCORE',
    'data__sampleMetrics__metrics__0__value': 1302,
    'data__sampleMetrics__metrics__0__UOM': 'NA',
    'data__sampleMetrics__metrics__1__demo': 11,
    'data__sampleMetrics__name': 'DNA Library QC Metrics',
    'data__sampleMetrics__passQualityControl': 'true'
}

如何通过将其索引视为键来拼合包含列表的字典?

推荐答案

您还需要检查列表-它们不是

You need to check for lists as well - they are not MutableMapings - hence they currently fall under your else: part and get added as is:

import collections
from itertools import chain 

def flatten(d, parent_key='', sep='__'):
    items = []
    for k, v in d.items():
        new_key = parent_key + sep + k if parent_key else k
        if isinstance(v, collections.MutableMapping):
            items.extend(flatten(v, new_key, sep=sep).items())
        elif isinstance(v, list):
            for idx, value in enumerate(v):
                items.extend(flatten(value, new_key + sep + str(idx), sep).items())
        else:
            items.append((new_key, v))
    return dict(items)

data = {
    "checksum": "c540fcd985bf88c87e48c2bfa1df5498",
    "data": {
        "sampleMetrics": {
            "name": "DNA Library QC Metrics",
            "passQualityControl": "true",
            "metrics": [{
                "name": "CONTAMINATION_SCORE",
                "value": 1302,
                "LSL": 0,
                "USL": 3106,
                "UOM": "NA"
            },{ 'demo': 11}]
        }
    }
}

print flatten(data) 

输出:

{'data__sampleMetrics__metrics__0__LSL': 0, 
 'checksum': 'c540fcd985bf88c87e48c2bfa1df5498', 
 'data__sampleMetrics__metrics__0__name': 'CONTAMINATION_SCORE', 
 'data__sampleMetrics__metrics__1__demo': 11, 
 'data__sampleMetrics__metrics__0__UOM': 'NA', 
 'data__sampleMetrics__metrics__0__USL': 3106, 
 'data__sampleMetrics__metrics__0__value': 1302, 
 'data__sampleMetrics__passQualityControl': 'true', 
 'data__sampleMetrics__name': 'DNA Library QC Metrics'}

---

要获取排序的"输出，您需要在python 2.x中使用OrderedDict:

---

To get a "sorted" output you would need to use an OrderedDict in python 2.x:

从集合中导入OrderedDict

from collections import OrderedDict

data = OrderedDict(sorted(flatten(data).items()))
print data 

输出:

OrderedDict([('checksum', 'c540fcd985bf88c87e48c2bfa1df5498'), 
             ('data__sampleMetrics__metrics__0__LSL', 0), 
             ('data__sampleMetrics__metrics__0__UOM', 'NA'), 
             ('data__sampleMetrics__metrics__0__USL', 3106), 
             ('data__sampleMetrics__metrics__0__name', 'CONTAMINATION_SCORE'), 
             ('data__sampleMetrics__metrics__0__value', 1302), 
             ('data__sampleMetrics__metrics__1__demo', 11), 
             ('data__sampleMetrics__name', 'DNA Library QC Metrics'), 
             ('data__sampleMetrics__passQualityControl', 'true')])

这篇关于如何通过将列表的索引视为键来使包含列表的字典扁平化?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！