python字典默认值

如何解决python字典默认值

有一个代码 - 它有效。但我不知道在这种情况下是否可以使用 'try-except'。

row = [('normal','Deck1','user1','Deck2','Win','2:0'),('normal','Deck3','user2','Loss','1:2'),'user3','Draw','2:2')]
table = 'win_loss'
result = ['Win','Draw']
deck = ['Deck1','Deck4']

def pivot(row):
  title,pos = [result,-2] if table == 'win_loss' else [deck,3]
  upd = {}
  for i in row:
    try:
      upd[i[1]]
    except KeyError:
      upd[i[1]] = {}
    try:
      upd[i[1]][i[pos]]
    except KeyError:
      upd[i[1]][i[pos]] = 0
    upd[i[1]][i[pos]] += 1
  return upd
  
print(pivot(row))
#{'Deck1': {'Win': 1},'Deck3': {'Loss': 1,'Draw': 1}}

有一个没有 'try-except' 的选项，但这会在字典的第二级留下空值。而且我不知道如何删除它们了。

def pivot(row):
  title,3]
  upd = {fr:{t:0 for t in title} for fr in deck}
  for i in row:
    upd[i[1]][i[pos]] += 1
#{'Deck1': {'Win': 1,'Loss': 0,'Draw': 0},'Deck2': {'Win': 0,'Deck3': {'Win': 0,'Loss': 1,'Draw': 1},'Deck4': {'Win': 0,'Draw': 0}}
  copy_upd = upd.copy()
  for key in copy_upd:
    if sum(copy_upd[key].values()) == 0:
      del upd[key]
  return upd
  #{'Deck1': {'Win': 1,'Draw': 1}}

print(pivot(row))

可以删除空值吗？或者如何为字典中没有的键设置默认值？

解决方法

感谢 Woodford 提供链接 Dictionaries and default values

def pivot(row):
  from collections import defaultdict
  title,pos = [result,-2] if table == 'win_loss' else [deck,3]
  upd = defaultdict(lambda: defaultdict(int))
  for i in row:
    upd[i[1]][i[pos]] += 1
  return upd
  #defaultdict(<function pivot.<locals>.<lambda> at 0x7f05740dc940>,{'Deck1': defaultdict(<class 'int'>,{'Win': 1}),'Deck3': defaultdict(<class 'int'>,{'Loss': 1,'Draw': 1})})