Question

使用to_dict（）我想出了以下字典。我需要放弃所有的纳米值。这种方法不起作用，因为它在迭代期间改变了大小。还有另一种方法可以实现这个目标吗？

{'k': {'a': nan, 'b': 1.0, 'c': 1.0},
 'u': {'a': 1.0, 'b': nan, 'c': 1.0},
 'y': {'a': nan, 'b': 1.0, 'c': nan}}
In [196]:

for listing, rate in master_dict.items():
for key in rate:
    if pd.isnull(master_dict[listing][key]) == True:
        del master_dict[listing][key]

---------------------------------------------------------------------------
RuntimeError                              Traceback (most recent call last)
<ipython-input-200-8859eb717bb9> in <module>()
      1 for listing, rate in master_dict.items():
----> 2     for key in rate:
      3         if pd.isnull(master_dict[listing][key]) == True:
      4             del master_dict[listing][key]
      5 

RuntimeError: dictionary changed size during iteration

Answer 1

您可以使用pandas.notnull或pandas.isnull进行过滤的dict comprehension加倍：

y = {k1:{k:v for k,v in v1.items() if pd.notnull(v)} for k1, v1 in d.items()}
print (y)
{'k': {'c': 1.0, 'b': 1.0}, 'u': {'c': 1.0, 'a': 1.0}, 'y': {'b': 1.0}}

类似的解决方案：

y = {k1:{k:v for k,v in v1.items() if not pd.isnull(v)} for k1, v1 in d.items()}
print (y)
{'k': {'c': 1.0, 'b': 1.0}, 'u': {'c': 1.0, 'a': 1.0}, 'y': {'b': 1.0}}

Answer 2

代码的增量修复将是：

for key in list(rate.keys()):  # make a copy and Python won't complain

更加pythonic的解决方案是使用dict compenhension删除项目：

for listing, rate in master_dict.items():
    master_dict[listing] = {k: v for k, v in rate.items() if pd.notnull(v)}

从字典中的字典中删除NaN（Dict在运行时错误期间更改大小）

2 个答案: