我没有做过很多Python编程，我试图在基本的csv中读取，然后从中创建一个嵌套的字典。这是我到目前为止所有，我似乎有一些循环或覆盖我的字典的问题。我知道它效率不高。
import csv
reader = csv.DictReader(open("fruit.csv"))
fruit_dict = {}
color_dict = {}
for row in reader:
info_list = []
count = row.pop('count')
info_list.append(count)
year = row.pop('year')
info_list.append(year)
info = row.pop('info')
info_list.append(info)
if row['color'] not in color_dict:
#print row['color']
color_dict['color'] = row['color']
#print fruit_dict
if row['fruit'] not in fruit_dict:
fruit_dict['name'] = row['fruit']
#print fruit_dict
#print info_list
list_of_info_lists =[]
list_of_info_lists.append(info_list)
fruit_dict['fruitInfo'] = list_of_info_lists
color_dict['fruit'] = fruit_dict
#print color_dict
else:
list_of_info_lists.append(info_list)
fruit_dict['fruitInfo'] = list_of_info_lists
color_dict['fruit'] = fruit_dict
#print color_dict
else:
if row['color'] in color_dict:
if row['fruit'] not in fruit_dict:
fruit_dict['name'] = row['fruit']
#print fruit_dict
#print info_list
list_of_info_lists =[]
list_of_info_lists.append(info_list)
fruit_dict['fruitInfo'] = list_of_info_lists
color_dict['fruit'] = fruit_dict
#print color_dict
else:
list_of_info_lists.append(info_list)
fruit_dict['fruitInfo'] = list_of_info_lists
color_dict['fruit'] = fruit_dict
#print color_dict
#print color_dict
这是csv：
color,fruit,year,count,info
red,apple,1970,3,good
red,apple,1922,5,okay
orange,orange,1935,2,okay
green,celery,2001,22,marginal
red,cherries,1999,5,outstanding
orange,carrot,1952,7,okay
green,celery,2014,2,good
green,grapes,2001,12,good
我得到的是：
{'color': 'green', 'fruit': {'name': 'grapes', 'fruitInfo': [['12', '2001', 'good']]}}
这很可爱，除了我期待比这更多的一行，并期待一个列表列表的名称＆＃39;已存在，例如：
{'color': 'red', 'fruit': {'name': 'apple', 'fruitInfo': [['5', '1922', 'okay'],['3', '1970', 'good']]}}
任何建议都将不胜感激。最终的目标是生成一个json文件。
谢谢， 苏珊
这是我最后想要的格式：
[{'color': 'red', 'fruit': {'name': 'apple', 'fruitInfo': [['5', '1922', 'okay'],['3', '1970', 'good']]}},
{'color': 'red', 'fruit': {'name': 'cherries', 'fruitInfo': [['5', '1999', 'outstanding']]}},
{'color': 'orange', 'fruit': {'name': 'orange', 'fruitInfo': [['2', '1935', 'okay']]}},
{'color': 'orange', 'fruit': {'name': 'carrot', 'fruitInfo': [['7', '1952', 'okay']]}},
{'color': 'green', 'fruit': {'name': 'celery', 'fruitInfo': [['2', '2014', 'good'],['22', '2001', 'marginal']]}},
{'color': 'green', 'fruit': {'name': 'grapes', 'fruitInfo': [['12', '2001', 'good']]}}]
答案 0 :(得分：2)
Jon Clements的回答是最佳解决方案。如果您想要了解最初开始帮助您了解可能出错的地方，请查看以下内容：
results_list = []
colorFruitTuple_set = set()
for row in reader:
info_list = [row['count'], row['year'],row['info']]
if (row['color'], row['fruit']) not in colorFruitTuple_set:
color_dict = {}
fruit_dict = {}
color_dict['color'] = row['color']
fruit_dict['name'] = row['fruit']
list_of_info_lists = [info_list]
fruit_dict['fruitInfo'] = list_of_info_lists
color_dict['fruit'] = fruit_dict
results_list.append(color_dict)
colorFruitTuple_set.add((row['color'], row['fruit']))
else:
for color_dict in results_list:
if color_dict["color"] == row['color'] and color_dict["fruit"]["name"] == row["fruit"]:
color_dict["fruit"]["fruitInfo"].append(info_list)
我认为这与你的目标一致。当您需要创建多个时，您尝试使用相同的color_dict和fruit_dict - 这也意味着您无法使用它们来跟踪重复项。这仅仅是为了学习目的 - Jon的方式是正确的方法。
希望这有帮助！
答案 1 :(得分：2)
您可以在此处使用defaultdict列表，将fruitInfo和2元组作为您的密钥（颜色和水果），然后重新格式化，例如：
import csv
from collections import defaultdict
dd = defaultdict(list)
with open('yourfile.csv') as fin:
csvin = csv.DictReader(fin)
for row in csvin:
dd[row['color'], row['fruit']].append([row['count'], row['year'], row['info']])
然后使用：
稍微重新格式化dd
reformatted = [{'color': c, 'fruit': {'name': f, 'fruitInfo': v}} for (c, f), v in dd.items()]
给你：
[{'color': 'orange',
'fruit': {'fruitInfo': [['7', '1952', 'okay']], 'name': 'carrot'}},
{'color': 'green',
'fruit': {'fruitInfo': [['12', '2001', 'good']], 'name': 'grapes'}},
{'color': 'orange',
'fruit': {'fruitInfo': [['2', '1935', 'okay']], 'name': 'orange'}},
{'color': 'red',
'fruit': {'fruitInfo': [['3', '1970', 'good'], ['5', '1922', 'okay']],
'name': 'apple'}},
{'color': 'red',
'fruit': {'fruitInfo': [['5', '1999', 'outstanding']], 'name': 'cherries'}},
{'color': 'green',
'fruit': {'fruitInfo': [['22', '2001', 'marginal'], ['2', '2014', 'good']],
'name': 'celery'}}]
答案 2 :(得分：0)
在处理字典词典时，我的模式是这样的：
sub_dict = main_dict.get(key, {})
sub_dict[sub_key] = sub_value
main_dict[key] = sub_dict
这会获取子词典，如果它不存在，则为{}。然后它为子字典赋值，并将子字典放回主字典中。
fruit_dict = {}
for row in reader:
# make the info_list
info_list = [row['count'], row['year'], row['info']]
# extract color and fruit into variables
color = row['color']
fruit = row['fruit']
# unpack the dictionaries and list
colors = fruit_dict.get(color, {})
fruits = colors.get(fruit, {})
info = fruits.get('info', [])
# reassemble the list and dictionaries
info.append(info_list)
fruits['info'] = info
colors[fruit] = fruits
fruit_dict[color] = colors
结果与您的示例略有不同，但需要更改它以使用颜色和水果作为键。
{＆＃39; orange＆＃39;：{＆＃39; orange＆＃39;：{＆＃39; info＆＃39;：[[＆＃39; 2＆＃39;，＆＃39; 1935＆＃ 39;，＆＃39;好的＆＃39;]]}，＆＃39;胡萝卜＆＃39;：{＆＃39; info＆＃39;：[[＆＃39; 7＆＃39;，＆＃39; 1952＆ ＃39;，＆＃39;好的＆＃39;]]}}，＆＃39;绿色＆＃39;：{＆＃39; celery＆＃39;：{＆＃39; info＆＃39;：[[＆＃ 39; 22＆＃39;，＆＃39; 2001＆＃39;，＆＃39; marginal＆＃39;]，[＆＃39; 2＆＃39;，＆＃39; 2014＆＃39;，＆＃39; good＆ ＃39;]]}，＆＃39; grape＆＃39;：{＆＃39; info＆＃39;：[[＆＃39; 12＆＃39;，＆＃39; 2001＆＃39;，＆＃39;好的＆＃39;]]}}，＆＃39; red＆＃39;：{＆＃39; cherries＆＃39;：{＆＃39; info＆＃39;：[[＆＃39; 5＆＃39;，＆ ＃39; 1999＆＃39;，＆＃39;杰出＆＃39;]]}，＆＃39; apple＆＃39;：{＆＃39; info＆＃39;：[[＆＃39; 3＆＃39;， ＆＃39; 1970＆＃39;，＆＃39; good＆＃39;]，[＆＃39; 5＆＃39;，＆＃39; 1922＆＃39;，＆＃39;好的＆＃39;]]}} }