问题:如何在Python中创建嵌套字典?
我有2个CSV文件:“数据”和“映射”:
- ‘映射’文件有4列:
Device_Name
,GDN
,Device_Type
,和Device_OS
。填充所有四个列。 - “数据”文件具有这些相同的列,其中
Device_Name
填充了列,而其他三列为空白。 - 我希望我的Python代码来打开这两个文件并为每个
Device_Name
数据文件,它的映射GDN
,Device_Type
以及Device_OS
从映射文件中值。
我知道只有2列存在时才需要使用dict(需要映射1列),但是当需要映射3列时我不知道如何实现。
以下是我尝试完成的映射的代码Device_Type
:
x = dict([])
with open("Pricing Mapping_2013-04-22.csv", "rb") as in_file1:
file_map = csv.reader(in_file1, delimiter=',')
for row in file_map:
typemap = [row[0],row[2]]
x.append(typemap)
with open("Pricing_Updated_Cleaned.csv", "rb") as in_file2, open("Data Scraper_GDN.csv", "wb") as out_file:
writer = csv.writer(out_file, delimiter=',')
for row in csv.reader(in_file2, delimiter=','):
try:
row[27] = x[row[11]]
except KeyError:
row[27] = ""
writer.writerow(row)
它返回Attribute Error
。
经过研究后,我认为我需要创建一个嵌套的字典,但是我不知道如何执行此操作。
回答 0
嵌套字典是字典中的字典。非常简单的事情。
>>> d = {}
>>> d['dict1'] = {}
>>> d['dict1']['innerkey'] = 'value'
>>> d
{'dict1': {'innerkey': 'value'}}
你也可以使用一个defaultdict
从collections
包装,以方便创建嵌套的字典。
>>> import collections
>>> d = collections.defaultdict(dict)
>>> d['dict1']['innerkey'] = 'value'
>>> d # currently a defaultdict type
defaultdict(<type 'dict'>, {'dict1': {'innerkey': 'value'}})
>>> dict(d) # but is exactly like a normal dictionary.
{'dict1': {'innerkey': 'value'}}
您可以根据需要填充。
我建议在你的代码的东西像下面:
d = {} # can use defaultdict(dict) instead
for row in file_map:
# derive row key from something
# when using defaultdict, we can skip the next step creating a dictionary on row_key
d[row_key] = {}
for idx, col in enumerate(row):
d[row_key][idx] = col
根据您的评论:
可能上面的代码令人困惑。我的问题简而言之:我有2个文件a.csv b.csv,a.csv有4列ijkl,b.csv也有这些列。我是这些csv的关键列。jkl列在a.csv中为空,但在b.csv中填充。我想使用’i’作为键列将b.csv中的jk l列的值映射到a.csv文件
我的建议是什么像这样(不使用defaultdict):
a_file = "path/to/a.csv"
b_file = "path/to/b.csv"
# read from file a.csv
with open(a_file) as f:
# skip headers
f.next()
# get first colum as keys
keys = (line.split(',')[0] for line in f)
# create empty dictionary:
d = {}
# read from file b.csv
with open(b_file) as f:
# gather headers except first key header
headers = f.next().split(',')[1:]
# iterate lines
for line in f:
# gather the colums
cols = line.strip().split(',')
# check to make sure this key should be mapped.
if cols[0] not in keys:
continue
# add key to dict
d[cols[0]] = dict(
# inner keys are the header names, values are columns
(headers[idx], v) for idx, v in enumerate(cols[1:]))
但是请注意,用于解析csv文件的是csv模块。
回答 1
更新:对于嵌套字典的任意长度,请转到此答案。
使用集合中的defaultdict函数。
高性能:当数据集很大时,“ if key not in dict”非常昂贵。
维护成本低:使代码更具可读性,并且可以轻松扩展。
from collections import defaultdict
target_dict = defaultdict(dict)
target_dict[key1][key2] = val
回答 2
对于任意级别的嵌套:
In [2]: def nested_dict():
...: return collections.defaultdict(nested_dict)
...:
In [3]: a = nested_dict()
In [4]: a
Out[4]: defaultdict(<function __main__.nested_dict>, {})
In [5]: a['a']['b']['c'] = 1
In [6]: a
Out[6]:
defaultdict(<function __main__.nested_dict>,
{'a': defaultdict(<function __main__.nested_dict>,
{'b': defaultdict(<function __main__.nested_dict>,
{'c': 1})})})
回答 3
重要的是要记住,在使用defaultdict和类似的嵌套dict模块(如nested_dict
)时,查找不存在的键可能会无意间在dict中创建新的键条目,并造成很多破坏。
这是带有nested_dict
模块的Python3示例:
import nested_dict as nd
nest = nd.nested_dict()
nest['outer1']['inner1'] = 'v11'
nest['outer1']['inner2'] = 'v12'
print('original nested dict: \n', nest)
try:
nest['outer1']['wrong_key1']
except KeyError as e:
print('exception missing key', e)
print('nested dict after lookup with missing key. no exception raised:\n', nest)
# Instead, convert back to normal dict...
nest_d = nest.to_dict(nest)
try:
print('converted to normal dict. Trying to lookup Wrong_key2')
nest_d['outer1']['wrong_key2']
except KeyError as e:
print('exception missing key', e)
else:
print(' no exception raised:\n')
# ...or use dict.keys to check if key in nested dict
print('checking with dict.keys')
print(list(nest['outer1'].keys()))
if 'wrong_key3' in list(nest.keys()):
print('found wrong_key3')
else:
print(' did not find wrong_key3')
输出为:
original nested dict: {"outer1": {"inner2": "v12", "inner1": "v11"}}
nested dict after lookup with missing key. no exception raised:
{"outer1": {"wrong_key1": {}, "inner2": "v12", "inner1": "v11"}}
converted to normal dict.
Trying to lookup Wrong_key2
exception missing key 'wrong_key2'
checking with dict.keys
['wrong_key1', 'inner2', 'inner1']
did not find wrong_key3