通过循环问题重命名字典键

Question

我在我的dataframe，df中有一列'EDU‘。在这里，我尝试用value_counts()，poe_dict创建一个字典。它看起来像这样。

edu_m=df['EDU'].sort_values()
poe_dict = edu_m.value_counts(normalize=True).to_dict()
poe_dict

{4: 0.47974705779026877,
 3: 0.24588090637625154,
 2: 0.172352011241876,
 1: 0.10202002459160373}

现在，我试图用这些字符串替换键'4,3,2,1‘，我把这些字符串放在一个列表中。

N_keys=“学院”，“比高中更多，但不是大学”，“高中”，“低于高中”

如果我每个人都这样做，运行正常，给我预期的结果。

In:
   poe_dict['college'] = poe_dict.pop(4)
   poe_dict['more than high school but not college'] = poe_dict.pop(3)
   poe_dict['high school'] = poe_dict.pop(2)
   poe_dict['less than high school'] = poe_dict.pop(1)

Out:
{'college': 0.47974705779026877,
'more than high school but not college': 0.24588090637625154,
'high school': 0.172352011241876,
'less than high school': 0.10202002459160373}

但是，如果我尝试将其作为一个循环来执行，它就会产生这样的结果。

In:
for key, n_key in zip(poe_dict.keys(), n_keys):
   poe_dict[n_key] = poe_dict.pop(key)
poe_dict

Out:
{2: 0.172352011241876,
1: 0.10202002459160373,
'high school': 0.47974705779026877,
'less than high school': 0.24588090637625154}

所以我不明白为什么这个循环不适用于键2和1？

我也尝试过调试它，以查看循环中会发生什么事情。

In:
for key, n_key in zip(poe_dict.keys(), n_keys):
  
   print (key,n_key)
   poe_dict[n_key] = poe_dict.pop(key)
   
Out:
4 college
3 more than high school but not college
college high school
more than high school but not college less than high school

Answer 1

在for循环中遍历poe_dict的键。但是，在运行poe_dict语句是poe_dict[n_key] = poe_dict.pop(key)时，会修改poe_dict[n_key] = poe_dict.pop(key)的键。因此，密钥信息会出错。正确的方法是将peo_dict的键存储到list list(poe_dict.keys())中，并循环这个新的键列表。

poe_dict = {4: 0.47, 3:0.25, 2:0.17, 1:0.10}

n_keys = ['college', 'more than high school but not college','high school', 'less than high school' ]
keylist = list(poe_dict.keys())
for key, n_key in zip(keylist, n_keys):
   print (key,n_key)
   poe_dict[n_key] = poe_dict.pop(key)
print (poe_dict)

结果将是

{'college': 0.47, 'more than high school but not college': 0.25, 'high school': 0.17, 'less than high school': 0.1}