字典串列如下
l = [
{'firstname': 'joe', 'surname': 'ABCD', 'tile' : 'DE'},
{'firstname': 'john', 'surname': 'DEF', 'tile' : 'BC'},
{'firstname': 'joe', 'surname': 'bloggs', 'tile' : 'DE'},
{'firstname': 'jane', 'surname': 'HH', 'tile' : 'AD'}
]
需要從liffirstname和tilematch中洗掉專案
偽代碼
for i in l:
if i['firstname'] i['tile'] in l:
l.pop(i)
我已經通過python 從串列中洗掉重復的字典
它洗掉整個字典匹配
還經歷了第二個答案python 從串列中洗掉重復的字典
uj5u.com熱心網友回復：
嘗試：
l = [
{"firstname": "joe", "surname": "ABCD", "tile": "DE"},
{"firstname": "john", "surname": "DEF", "tile": "BC"},
{"firstname": "joe", "surname": "bloggs", "tile": "DE"},
{"firstname": "jane", "surname": "HH", "tile": "AD"},
]
out = {}
for d in reversed(l):
out[(d["firstname"], d["tile"])] = d
print(list(out.values()))
印刷：
[
{"firstname": "jane", "surname": "HH", "tile": "AD"},
{"firstname": "joe", "surname": "ABCD", "tile": "DE"},
{"firstname": "john", "surname": "DEF", "tile": "BC"},
]
uj5u.com熱心網友回復：
試試這個，
擴展了這個 SO answer Index of duplicates items in a python list
代碼：
from collections import defaultdict
l = [
{'firstname': 'joe', 'surname': 'ABCD', 'tile' : 'DE'},
{'firstname': 'john', 'surname': 'DEF', 'tile' : 'BC'},
{'firstname': 'joe', 'surname': 'bloggs', 'tile' : 'DE'},
{'firstname': 'jane', 'surname': 'HH', 'tile' : 'AD'}
]
group = [(item['firstname'], item['tile']) for item in l]
def list_duplicates(seq):
tally = defaultdict(list)
for i,item in enumerate(seq):
tally[item].append(i)
return ((key,locs) for key,locs in tally.items()
if len(locs)>=1)
new_l = []
for dup in list_duplicates(group):
new_l.append(l[dup[1][0]])
new_l
輸出：
[{'firstname': 'joe', 'surname': 'ABCD', 'tile': 'DE'},
{'firstname': 'john', 'surname': 'DEF', 'tile': 'BC'},
{'firstname': 'jane', 'surname': 'HH', 'tile': 'AD'}]
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