[英]python itertools groupby return tuple
我需要解析扁平化的結構並使用提供的鍵列表創建嵌套的結構。 我已經解決了問題,但是我正在尋求改進,我想學習可以在代碼中進行哪些更改。 有人可以復習並使用更好的知識進行重構嗎?
src_data = [
{
"key1": "XX",
"key2": "X111",
"key3": "1aa",
"key4": 1
},
{
"key1": "YY",
"key2": "Y111",
"key3": "1bb",
"key4": 11
},
{
"key1": "ZZ",
"key2": "Z111",
"key3": "1cc",
"key4": 2.4
},
{
"key1": "AA",
"key2": "A111",
"key3": "1cc",
"key4": 33333.2122
},
{
"key1": "BB",
"key2": "B111",
"key3": "1bb",
"key4": 2
},
]
這是我到目前為止開發的代碼,用於創建最終結果。
def plant_tree(ll):
master_tree = {}
for i in ll:
tree = master_tree
for n in i:
if n not in tree:
tree[n] = {}
tree = tree[n]
return master_tree
def make_nested_object(tt, var):
elo = lambda l: reduce(lambda x, y: {y: x}, l[::-1], var)
return {'n_path': tt, 'n_structure': elo(tt)}
def getFromDict(dataDict, mapList):
return reduce(operator.getitem, mapList, dataDict)
def set_nested_item(dataDict, mapList, val):
"""Set item in nested dictionary"""
reduce(getitem, mapList[:-1], dataDict)[mapList[-1]] = val
return dataDict
def update_tree(data_tree):
# MAKE NESTED OBJECT
out = (make_nested_object(k, v) for k,v, in res_out.items())
for dd in out:
leaf_data = dd['n_structure']
leaf_path = dd['n_path']
data_tree = set_nested_item(data_tree, leaf_path, getFromDict(leaf_data, leaf_path))
return data_tree
這是此問題中的自定義itemgeter函數
def customed_itemgetter(*args):
# this handles the case when one key is provided
f = itemgetter(*args)
if len(args) > 2:
return f
return lambda obj: (f(obj),)
定義嵌套級別
nesting_keys = ['key1', 'key3', 'key2']
grouper = customed_itemgetter(*nesting_keys)
ii = groupby(sorted(src_data, key=grouper), grouper)
res_out = {key: [{k:v for k,v in i.items() if k not in nesting_keys} for i in group] for key,group in ii}
#
ll = ([dd[x] for x in nesting_keys] for dd in src_data)
data_tree = plant_tree(ll)
得到結果
result = update_tree(data_tree)
如何改善我的代碼?
如果itemgetter
[Python的DOC]中給出一個單一的元件,它返回單個元件,並且在一個單元組不包裹。
但是,我們可以為此構建一個函數,例如:
from operator import itemgetter
def itemgetter2(*args):
f = itemgetter(*args)
if len(args) > 2:
return f
return lambda obj: (f(obj),)
然后我們可以使用新的itemgetter2
,例如:
grouper = itemgetter2(*ll)
ii = groupby(sorted(src_data, key=grouper), grouper)
編輯 :但是,根據您的問題,您想要執行多級分組,我們可以為此創建一個函數,例如:
def multigroup(groups, iterable, index=0):
if len(groups) <= index:
return list(iterable)
else:
f = itemgetter(groups[index])
i1 = index + 1
return {
k: multigroup(groups, vs, index=i1)
for k, vs in groupby(sorted(iterable, key=f), f)
}
對於問題中的data_src
,這將生成:
>>> multigroup(['a', 'b'], src_data)
{1: {2: [{'a': 1, 'b': 2, 'z': 3}]}, 2: {3: [{'a': 2, 'b': 3, 'e': 2}]}, 4: {3: [{'a': 4, 'x': 3, 'b': 3}]}}
但是,您可以對list(..)
調用中的值進行后處理。 例如,我們可以生成沒有分組列中的元素的字典:
def multigroup(groups, iterable):
group_set = set(groups)
fs = [itemgetter(group) for group in groups]
def mg(iterable, index=0):
if len(groups) <= index:
return [
{k: v for k, v in item.items() if k not in group_set}
for item in iterable
]
else:
i1 = index + 1
return {
k: mg(vs, index=i1)
for k, vs in groupby(sorted(iterable, key=fs[index]), fs[index])
}
return mg(iterable)
對於給定的樣本輸入,我們得到:
>>> multigroup(['a', 'b'], src_data)
{1: {2: [{'z': 3}]}, 2: {3: [{'e': 2}]}, 4: {3: [{'x': 3}]}}
或對於新的樣本數據:
>>> pprint(multigroup(['key1', 'key3', 'key2'], src_data))
{'AA': {'1cc': {'A111': [{'key4': 33333.2122}]}},
'BB': {'1bb': {'B111': [{'key4': 2}]}},
'XX': {'1aa': {'X111': [{'key4': 1}]}},
'YY': {'1bb': {'Y111': [{'key4': 11}]}},
'ZZ': {'1cc': {'Z111': [{'key4': 2.4}]}}}
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