[英]Create keys for dictionary based on length of list of lists
我有以下列表:
my_list = [[['pd', 1],
['pd_de', None],
['pd_amnt', '$10.00']],
[['pd', 1],
['pd_de', '5/1/19 '],
['pd_amnt', '$100.00 '],
['pd', 2],
['pd_de', '5/1/20 '],
['pd_amnt', '$200.00 ']],
[['pd', 1],
['pd_de', None],
['pd_amnt', None]],
[['pd', 1],
['pd_de', '5/1/19 '],
['pd_amnt', '$300.00 '],
['pd', 2],
['pd_de', '5/1/20 '],
['pd_amnt', '$600.00 '],
['pd', 3],
['pd_de', '6/1/18'],
['pd_amnt', '$450.00']]]
使用它,我想创建一个字典列表。 我在下面创建字典列表,
list_dict = []
for i in my_list:
temp_dict = {}
for j in i:
temp_dict[j[0]] = j[1]
list_dict.append(temp_dict)
我得到一个像这样的 output,我不想要,
[{'pd': 1, 'pd_de': None, 'pd_amnt': '$10.00'},
{'pd': 2, 'pd_de': '5/1/20 ', 'pd_amnt': '$200.00 '},
{'pd': 1, 'pd_de': None, 'pd_amnt': None},
{'pd': 3, 'pd_de': '6/1/18', 'pd_amnt': '$450.00'}]
我需要这样的 output,
[{'pd_1': 1, 'pd_de_1': None, 'pd_amnt_1': '$10.00'},
{'pd_1': 1, 'pd_de_1': '5/1/19', 'pd_amnt_1': '$100.00', 'pd_2': 2, 'pd_de_2': '5/1/20 ', 'pd_amnt_2': '$200.00 '},
{'pd_1': 1, 'pd_de_1': None, 'pd_amnt_1': None},
{'pd_1': 1, 'pd_de_1': '5/1/19', 'pd_amnt_1': '$300.00','pd_2': 2, 'pd_de_2': '5/1/20', 'pd_amnt': '$600.00','pd_3': 1, 'pd_de_3': '6/1/18', 'pd_amnt_3': '$450.00'}]
如果你在上面看到,当里面的列表长度为 3 时,它们是可以的。如果超过 3,那么它不会给出正确的结果。
当我为字典创建键时,我也不确定如何在键(即'pd_1')中创建"_"
。
如何实现我想要的 output?
(注意:不知道如何命名标题,我说的是列表长度,我可能错了,因为我不熟悉pythonic术语)
保留项目的顺序:
import pandas as pd
from collections import OrderedDict
# my_list = ...
res = []
for l1 in my_list:
d = OrderedDict()
for l2 in l1:
if l2[0] == 'pd':
sfx = l2[1]
d[f'{l2[0]}_{sfx}'] = l2[1].strip() if isinstance(l2[1], str) else l2[1]
res.append(d)
df = pd.DataFrame(res)
print(df)
output:
pd_1 pd_de_1 pd_amnt_1 pd_2 pd_de_2 pd_amnt_2 pd_3 pd_de_3 pd_amnt_3
0 1 None $10.00 NaN NaN NaN NaN NaN NaN
1 1 5/1/19 $100.00 2.0 5/1/20 $200.00 NaN NaN NaN
2 1 None None NaN NaN NaN NaN NaN NaN
3 1 5/1/19 $300.00 2.0 5/1/20 $600.00 3.0 6/1/18 $450.00
您可以使用附加变量( counter
)来查找字典中尚不存在的键“索引”:
result = []
for sub_list in my_list:
temp = {}
for key, value in sub_list:
counter = 1
while f"{key}_{counter}" in temp:
counter += 1
temp[f"{key}_{counter}"] = value
result.append(temp)
更有效的解决方案是将计数器存储到 dict 中,并在使用键后递增它们:
result = []
for sub_list in my_list:
counters = {}
temp = {}
for key, value in sub_list:
if key in counters:
counters[key] += 1
else:
counters[key] = 1
temp[f"{key}_{counters[key]}" ] = value
result.append(temp)
使用collections.defaultdict
你可以把它写得更短一点:
from collections import defaultdict
result = []
for sub_list in my_list:
counters = defaultdict(int)
temp = {}
for key, value in sub_list:
counters[key] += 1
temp[f"{key}_{counters[key]}"] = value
result.append(temp)
defaultdict
来增加键。 然后将其添加到您的result
字典中。list_dict = []
from collections import defaultdict
for i in my_list:
temp_dict = {}
incr = defaultdict(int)
for j in i:
incr[j[0]] += 1
temp_dict[j[0] + '_' + str(incr[j[0]])] = j[1]
list_dict.append(temp_dict)
Output:
[{'pd_1': 1, 'pd_de_1': None, 'pd_amnt_1': '$10.00'},
{'pd_1': 1,
'pd_de_1': '5/1/19 ',
'pd_amnt_1': '$100.00 ',
'pd_2': 2,
'pd_de_2': '5/1/20 ',
'pd_amnt_2': '$200.00 '},
{'pd_1': 1, 'pd_de_1': None, 'pd_amnt_1': None},
{'pd_1': 1,
'pd_de_1': '5/1/19 ',
'pd_amnt_1': '$300.00 ',
'pd_2': 2,
'pd_de_2': '5/1/20 ',
'pd_amnt_2': '$600.00 ',
'pd_3': 1,
'pd_de_3': '6/1/18',
'pd_amnt_3': '$450.00'}]
你得到这个的原因是当你在字典中设置一个键时,它会覆盖任何以前的数据。 例如,你有这个字典x = ["a":1, "b":2, "c":3]
如果你做x["d"] = 4
它将是["a":1, "b":2, "c":3, "d":4]
但如果你再做x["a"] = 3
它将是["a":3, "b":2, "c":3, "d":4]
。
您的解决方案是将每个项目添加到字典中,标签后带有一个数字来表示它是哪个标签。
list_dict = []
for i in my_list:
temp_dict = {}
for j in i:
a = 1
while j[0]+"_"+str(a) in temp_dict:
a += 1
temp_dict[j[0]+"_"+str(a)] = j[1]
list_dict.append(temp_dict)
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