我有兩個(gè)清單：第一個(gè)列表包含頂級(jí)名稱(chēng)信息；每個(gè)項(xiàng)目有多個(gè)描述符第二個(gè)列表包含第一個(gè)列表中項(xiàng)目的變體；每個(gè)項(xiàng)目還有多個(gè)描述符外部（第一個(gè)）列表包含三項(xiàng)；name1、name2、 和name3。還有一個(gè)source，但我沒(méi)有費(fèi)心在示例中更改它以保持簡(jiǎn)單。內(nèi)部（第二個(gè)）列表包含外部列表的名稱(chēng)和來(lái)源，然后有自己的名稱(chēng)和來(lái)源。最終結(jié)果應(yīng)該是這樣的：name1 (source1)===============* [name1-foo1]: . "description1"* [name1-foo2]: . "description2"* [name1-foo3]: . "description3"subname1 (subsource1)---------------* [name1-sub1-bar1]: . "description1"* [name1-sub1-bar2]: . "description2"* [name1-sub1-bar3]: . "description3"subname2 (subsource1)---------------* [name1-sub2-bar1]: . "description1"* [name1-sub2-bar2]: . "description2"* [name1-sub2-bar3]: . "description3"...我的問(wèn)題是我的外循環(huán)查找name和中的更改source作為打印標(biāo)題并繼續(xù)打印下一個(gè)內(nèi)容的觸發(fā)器。但由于內(nèi)循環(huán)僅由更改觸發(fā)，因此當(dāng)外循環(huán)運(yùn)行完時(shí)，它不會(huì)最后一次運(yùn)行內(nèi)循環(huán)來(lái)獲取所有子項(xiàng)。import collectionsOuterRecord = collections.namedtuple('OuterRecord', 'name, source, thing, level, description')InnerRecord = collections.namedtuple('InnerRecord', 'name, source, in_name, in_source, thing, level, description')o = [ OuterRecord('name1', 'source1', 'name1-foo1', 1, 'description1'), OuterRecord('name1', 'source1', 'name1-foo2', 5, 'description2'), OuterRecord('name1', 'source1', 'name1-foo3', 10, 'description3'), OuterRecord('name2', 'source1', 'name2-foo1', 1, 'description1'), OuterRecord('name2', 'source1', 'name2-foo2', 5, 'description2'), OuterRecord('name2', 'source1', 'name2-foo3', 10, 'description3'), OuterRecord('name3', 'source1', 'name3-foo1', 1, 'description1'), OuterRecord('name3', 'source1', 'name3-foo2', 5, 'description2')]]因此，一旦一切完成，我可以再次單獨(dú)運(yùn)行內(nèi)部循環(huán)，但這聞起來(lái)很糟糕。有沒(méi)有更好的方法來(lái)構(gòu)建循環(huán)，以便這一切一次性發(fā)生？
2 回答
jeck貓
TA貢獻(xiàn)1909條經(jīng)驗(yàn) 獲得超7個(gè)贊
您的代碼的一個(gè)有趣的事實(shí)是，它看起來(lái)具有檢測(cè)更改背后的價(jià)值。不幸的是，最后一個(gè)值沒(méi)有機(jī)會(huì)被檢測(cè)到，因?yàn)檠h(huán)在那里結(jié)束。
相反，使用兩個(gè)指針，一個(gè)用于當(dāng)前值，一個(gè)用于列表中的下一個(gè)值。使用這樣的條件來(lái)檢測(cè)下一個(gè)的變化（lookahead）
for index, outer in enumerate(outer_list):
next_outer_source = outer_list[index + 1] if index < len(outer_list) - 1 else None
show_outer_header = current_outer_name != outer.name or current_outer_source != outer.source
show_inner_values = next_outer_source is None or outer.name != next_outer_source.name or outer.source != next_outer_source.source
這是您的函數(shù)的清理副本：
def loop_over(outer_list, inner_list):
current_outer_name = None
current_outer_source = None
current_inner_name = None
current_inner_source = None
prev_outer_source = None
for index, outer in enumerate(outer_list):
next_outer_source = outer_list[index + 1] if index < len(outer_list) - 1 else None
show_outer_header = current_outer_name != outer.name or current_outer_source != outer.source
show_inner_values = next_outer_source is None or outer.name != next_outer_source.name or outer.source != next_outer_source.source
# print outer header
if show_outer_header:
print('\n{} ({})'.format(outer.name, outer.source))
print('=' * 15)
current_outer_name, current_outer_source = outer.name, outer.source
# print outer value
print('* [{}]: . "{}"'.format(outer.thing, outer.description))
# print inner values
if show_inner_values:
current_outer_name = outer.name
current_outer_source = outer.source
for inner in [x for x in inner_list if x.name == current_outer_name and x.source == current_outer_source]:
if current_inner_name is None:
print('\n{} ({})'.format(inner.in_name, inner.in_source))
print('-' * 15)
current_inner_name = inner.in_name
current_inner_source = inner.in_source
if inner.in_name != current_inner_name or inner.in_source != current_inner_source:
print('\n{} ({})'.format(inner.in_name, inner.in_source))
print('-' * 15)
current_inner_name = inner.in_name
current_inner_source = inner.in_source
print('* [{}]: . "{}"'.format(inner.thing, inner.description))
輸出：
name1 (source1)
===============
* [name1-foo1]: . "description1"
* [name1-foo2]: . "description2"
* [name1-foo3]: . "description3"
subname1 (subsource1)
---------------
* [name1-sub1-bar1]: . "description1"
* [name1-sub1-bar2]: . "description2"
* [name1-sub1-bar3]: . "description3"
subname2 (subsource1)
---------------
* [name1-sub2-bar1]: . "description1"
* [name1-sub2-bar2]: . "description2"
* [name1-sub2-bar3]: . "description3"
name2 (source1)
===============
* [name2-foo1]: . "description1"
* [name2-foo2]: . "description2"
* [name2-foo3]: . "description3"
subname3 (subsource1)
---------------
* [name2-sub3-bar1]: . "description1"
* [name2-sub3-bar2]: . "description2"
* [name2-sub3-bar2]: . "description3"
name3 (source1)
===============
* [name3-foo1]: . "description1"
* [name3-foo2]: . "description2"
subname4 (subsource1)
---------------
* [name3-sub4-bar1]: . "description1"
* [name3-sub4-bar2]: . "description2"
* [name3-sub4-bar2]: . "description3"
subname5 (subsource1)
---------------
* [name3-sub5-bar1]: . "description1"
* [name3-sub5-bar2]: . "description2"
* [name3-sub5-bar3]: . "description3"
subname6 (subsource1)
---------------
* [name3-sub6-bar1]: . "description1"
* [name3-sub6-bar2]: . "description2"
* [name3-sub6-bar3]: . "description3"
慕標(biāo)琳琳
TA貢獻(xiàn)1830條經(jīng)驗(yàn) 獲得超9個(gè)贊
考慮到詳細(xì)的數(shù)據(jù)結(jié)構(gòu)，我預(yù)計(jì)效率并不是您真正追求的，因此這似乎是一種相當(dāng)簡(jiǎn)潔且可讀的方式來(lái)獲取您所需要的內(nèi)容：
def print_outer_and_inner(outer_recs, inner_recs):
for name, source in {(o_rec.name, o_rec.source): None for o_rec in outer_recs}:
print(f'{name} ({source})')
print('=' * 15)
for o_rec in outer_recs:
if o_rec.name == name:
print(f'* [{o_rec.thing}]: . "{o_rec.description}"')
print()
for sub_name, sub_source in {(i_rec.in_name, i_rec.in_source): None for i_rec in inner_recs}:
print(f'{sub_name} ({sub_source})')
print('-' * 15)
for i_rec in inner_recs:
if i_rec.in_name == sub_name:
print(f'* [{i_rec.thing}]: . "{i_rec.description}"')
print()
print_outer_and_inner(o, i)
主要缺點(diǎn)是它會(huì)在每個(gè)列表上循環(huán)多次，但它給你帶來(lái)的是簡(jiǎn)潔性和可讀性。
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