Question

我想根据以下规则清理一系列的词组：

1）dicts列表已经排序，因此较早的dicts是首选 2）在较低的dicts中，如果['name']和['code']字符串值与列表中较高的任何dict的相同键值匹配，并且{{1的差值的绝对值这两个dicts之间是int(['cost']);然后该dict被假定为早期字典的副本，并从列表中删除。

以下是来自词典列表的一个词典：

< 2

删除像这样的重复项的最佳方法是什么？

Answer 1

这可能有更多的pythonic方式，但这是基本的伪代码：

def is_duplicate(a,b):
  if a['name'] == b['name'] and a['cost'] == b['cost'] and abs(int(a['cost']-b['cost'])) < 2:
    return True
  return False

newlist = []
for a in oldlist:
  isdupe = False
  for b in newlist:
    if is_duplicate(a,b):
      isdupe = True
      break
  if not isdupe:
    newlist.append(a)

Answer 2

由于您说费用是整数，您可以使用：

def neardup( items ):
    forbidden = set()
    for elem in items:
        key = elem['name'], elem['code'], int(elem['cost'])
        if key not in forbidden:
            yield elem
            for diff in (-1,0,1): # add all keys invalidated by this
                key = elem['name'], elem['code'], int(elem['cost'])-diff
                forbidden.add(key)

这是一种真正计算差异的不太棘手的方法：

from collections import defaultdict
def neardup2( items ):
    # this is a mapping `(name, code) -> [cost1, cost2, ... ]`
    forbidden =  defaultdict(list)
    for elem in items:
        key = elem['name'], elem['code']
        curcost = float(elem['cost'])
        # a item is new if we never saw the key before
        if (key not in forbidden or
              # or if all the known costs differ by more than 2
              all(abs(cost-curcost) >= 2 for cost in forbidden[key])):
            yield elem
            forbidden[key].append(curcost)

两种解决方案都避免重新扫描每个项目的整个列表。毕竟，如果(name, code)相等，成本才会变得有趣，因此您可以使用字典快速查找所有候选人。

Answer 3

有点令人费解的问题，但我认为这样的事情会起作用：

for i, d in enumerate(dictList):
    # iterate through the list of dicts, starting with the first
    for k,v in d.iteritems():
        # for each key-value pair in this dict...
        for d2 in dictList[i:]:
             # check against all of the other dicts "beneath" it
             # eg,
             # if d['name'] == d2['name'] and d['code'] == d2['code']:
             #     --check the cost stuff here--

从dicts列表中删除具有接近重复值的dicts - Python

3 个答案: