[英]Python Remove duplicates from list of dictionaries based on a value
我有字典列表
vals = [
{'tmpl_id': 67, 'qty_available': -3.0, 'product_id': 72, 'product_qty': 1.0},
{'tmpl_id': 67, 'qty_available': 5.0, 'product_id': 71, 'product_qty': 1.0}
{'tmpl_id': 69, 'qty_available': 10.0, 'product_id': 74, 'product_qty': 1.0}
]
from operator import itemgetter
getvals = operator.itemgetter('tmpl_id')
val.sort(key=getvals)
result = []
for k, g in itertools.groupby(val, getvals):
result.append(g.next())
val[:] = result
我想刪除重復值 (tmpl_id) 並且還基於 qty_available 較小或為負
輸出將類似於:
vals = [
{'tmpl_id': 67, 'qty_available': 5.0, 'product_id': 71, 'product_qty': 1.0}
{'tmpl_id': 69, 'qty_available': 10.0, 'product_id': 74, 'product_qty': 1.0}
]
from collections import Counter
vals = [{'tmpl_id': 67, 'qty_available': -3.0, 'product_id': 72, 'product_qty': 1.0},
{'tmpl_id': 67, 'qty_available': 5.0, 'product_id': 71, 'product_qty': 1.0},
{'tmpl_id': 69, 'qty_available': 10.0, 'product_id': 74, 'product_qty': 1.0},]
k = [x['tmpl_id'] for x in vals]
new_vals=[]
for i in Counter(k):
all = [x for x in vals if x['tmpl_id']==i]
new_vals.append(max(all, key=lambda x: x['qty_available']))
>>> new_vals
[
{'product_qty': 1.0, 'qty_available': 5.0, 'tmpl_id': 67, 'product_id': 71},
{'product_qty': 1.0, 'qty_available': 10.0, 'tmpl_id': 69, 'product_id': 74}
]
您可以使用"tmpl_id"
的值作為鍵來存儲"tmpl_id"
,將 dict 設置為值,如果您得到一個'qty_available'
更高的 dict,那么您將替換為當前的 dict :
def remove_dupes(l, k, k2):
seen = {}
for d in vals:
v, v2 = d[k], d[k2]
if v not in seen:
seen[v] = d
elif v2 > seen[v][k2]:
seen[v] = d
return seen
vals[:] = remove_dupes(vals, "tmpl_id",'qty_available' ).values()
輸出:
[{'product_id': 71, 'qty_available': 5.0, 'tmpl_id': 67, 'product_qty': 1.0},
{'product_id': 74, 'qty_available': 10.0, 'tmpl_id': 69, 'product_qty': 1.0}]
如果您要使用 sorted 和 groupby,您只需要反向排序並從每個 v 中獲取第一個值:
from itertools import groupby
from operator import itemgetter
keys = itemgetter("tmpl_id",'qty_available')
vals[:] = (next(v) for k,v in groupby(sorted(vals, key=keys,reverse=True),
key=itemgetter("tmpl_id")))
print(vals)
反轉排序意味着較高的'qty_available'
將首先出現,因此對於唯一的字典,它只會為您提供該字典,對於重復的 tmpl_id,您將獲得具有最大'qty_available'
值的那個。
如果您想要就地排序而不是創建新列表,只需使用vals.sort()
並刪除對 sorted 的調用
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