有没有更快的方法？

Question

所以我在做一个python项目，我需要按字典组织一个值列表。我想知道是否有比做我正在做的事情更快的方法。你知道吗

这就是我所做的，有没有更有效更简单的方法？你知道吗

def catogorize_by_mortality(hurricanes):
  damage_scale = {0: 0, 1: 100, 2: 500, 3: 1000, 4: 1e4}
  hurricane_mortality_dict = {0:[], 1:[], 2:[], 3:[], 4:[], 5:[]}
  for hurricane in hurricanes:
    current_hurricane = hurricanes[hurricane]
    death_count = current_hurricane['Deaths']
    if death_count > damage_scale[0] and death_count < damage_scale[1]:
      hurricane_mortality_dict[0] += hurricane

    elif death_count > damage_scale[1] and death_count < damage_scale[2]:
      hurricane_mortality_dict[1] += hurricane

    elif death_count > damage_scale[2] and death_count < damage_scale[3]:
      hurricane_mortality_dict[2] += hurricane

    elif death_count > damage_scale[3] and death_count < damage_scale[4]:
      hurricane_mortality_dict[3] += hurricane

    elif death_count >= damage_scale[4]:
      hurricane_mortality_dict[4] += hurricane

    else:
      hurricane_mortality_dict[5] += hurricane

  return hurricane_mortality_dict

# example of the hurricanes dictionary when printed
{'Cuba I': {'Name': 'Cuba I', 'Month': 'October', 'Year': 1924, 'Max Sustained Wind': 165, 'Areas Affected': ['Central America', 'Mexico', 'Cuba', 'Florida', 'The Bahamas'], 'Deaths': 90}

# this is what it returns
deaths.')
{0: ['C', 'u', 'b', 'a'

应该发生的是，它将返回飓风的名称分类，但相反，它分裂成字符他们发生了什么？你知道吗

Answer 1

hurricane_mortality_dict[i]定义为一个列表，位于函数顶部附近：

hurricane_mortality_dict = {0:[], 1:[], 2:[], 3:[], 4:[], 5:[]}

所以在条件语句中，当您执行hurricane_mortality_dict[i] += hurricane时，您试图添加一个string和一个list，它将字符串视为一个list字符-因此您的输出。你知道吗

您只需更改以下各项：

hurricane_mortality_dict[i] += hurricane

收件人：

hurricane_mortality_dict[i].append(hurricane)

对于您的示例输入，这将导致：

{0: ['Cuba I'], 1: [], 2: [], 3: [], 4: [], 5: []}