我已经将数据存储到excel列表中,但是现在我想将其存储到字典中。 here is the excel
在列表中给出这样的输出。
[['Nodeb_IN_New', 107, 'class-default', 'mobility-platinum', 'h1', 7, 'dscp-fc-map', ['ef']], ['', 107, '', 'mobility-gold-new', 'h2', 5, 'dscp-fc-map', ['af41']], ['', 107, '', 'mobility-silver-new', 'l1', 4, 'dscp-fc-map', ['af11', 'af21', 'af31']], ['Nokia_SRAN_S1-MME_X2_IN', 102, '', 'Nokia_SRAN_mobility_platinum', 'h1', 7, 'dscp-fc-map', ['ef', 'nc1']]]
我想和字典一样。我已经写了这段代码
def myfun(list):
list = list
res=dict()
qos=None
for row in list:
if row[0]=="":
res[qos]['b']=row[1]
res[qos]['c']=row[2]
res[qos]['d']=row[3]
res[qos]['e']=row[4]
res[qos]['f']=row[5]
res[qos]['g']=row[6]
res[qos]['h']=row[7]
else:
qos=row[0]
res[qos]=dict()
res[qos]['b']=row[1]
res[qos]['c']=row[2]
res[qos]['d']=row[3]
res[qos]['e']=row[4]
res[qos]['f']=row[5]
res[qos]['g']=row[6]
res[qos]['h']=row[7]
但是它给出了这样的输出。
{'Nodeb_IN_New': {
'b': 107, 'c': '', 'd': 'mobility-silver-new',
'e': 'l1', 'f': 4, 'g': 'dscp-fc-map',
'h': ['af11', 'af21', 'af31']
},
'Nokia_SRAN_S1-MME_X2_IN': {
'b': 102, 'c': '', 'd': 'Nokia_SRAN_mobility_platinum', 'e': 'h1', 'f': 7, 'g': 'dscp-fc-map', 'h': ['ef', 'nc1']}}
在“ Nodeb_IN_New”下,我希望所有3行(来自excel)。
答案 0 :(得分:1)
尝试此代码:
import json
l=[['Nodeb_IN_New', 107, 'class-default', 'mobility-platinum', 'h1', 7, 'dscp-fc-map', ['ef']], ['', 107, '', 'mobility-gold-new', 'h2', 5, 'dscp-fc-map', ['af41']], ['', 107, '', 'mobility-silver-new', 'l1', 4, 'dscp-fc-map', ['af11', 'af21', 'af31']], ['Nokia_SRAN_S1-MME_X2_IN', 102, '', 'Nokia_SRAN_mobility_platinum', 'h1', 7, 'dscp-fc-map', ['ef', 'nc1']]]
def myfun(list):
res=dict()
qos=None
d={}
li=[]
for row in list:
if row[0]=="":
res[qos]=dict()
res[qos]['b']=row[1]
res[qos]['c']=row[2]
res[qos]['d']=row[3]
res[qos]['e']=row[4]
res[qos]['f']=row[5]
res[qos]['g']=row[6]
res[qos]['h']=row[7]
else:
qos=row[0]
res[qos]=dict()
res[qos]['b']=row[1]
res[qos]['c']=row[2]
res[qos]['d']=row[3]
res[qos]['e']=row[4]
res[qos]['f']=row[5]
res[qos]['g']=row[6]
res[qos]['h']=row[7]
x = res.keys()
keylist = []
keylist.extend(iter(x))
if keylist[0] in d.keys():
d[keylist[0]].append(res[qos])
else:
d[keylist[0]] = []
d[keylist[0]].append(res[qos])
print(d)
print(json.dumps(d))
myfun(l)
输出:
{'Nokia_SRAN_S1-MME_X2_IN': [{'d': 'Nokia_SRAN_mobility_platinum', 'f': 7, 'e': 'h1', 'b': 102, 'c': '', 'h': ['ef', 'nc1'], 'g': 'dscp-fc-map'}], 'Nodeb_IN_New': [{'d': 'mobility-platinum', 'f': 7, 'e': 'h1', 'b': 107, 'c': 'class-default', 'h': ['ef'], 'g': 'dscp-fc-map'}, {'d': 'mobility-gold-new', 'f': 5, 'e': 'h2', 'b': 107, 'c': '', 'h': ['af41'], 'g': 'dscp-fc-map'}, {'d': 'mobility-silver-new', 'f': 4, 'e': 'l1', 'b': 107, 'c': '', 'h': ['af11', 'af21', 'af31'], 'g': 'dscp-fc-map'}]}
导致json可读:
{
"Nokia_SRAN_S1-MME_X2_IN": [
{
"d": "Nokia_SRAN_mobility_platinum",
"f": 7,
"e": "h1",
"b": 102,
"c": "",
"h": [
"ef",
"nc1"
],
"g": "dscp-fc-map"
}
],
"Nodeb_IN_New": [
{
"d": "mobility-platinum",
"f": 7,
"e": "h1",
"b": 107,
"c": "class-default",
"h": [
"ef"
],
"g": "dscp-fc-map"
},
{
"d": "mobility-gold-new",
"f": 5,
"e": "h2",
"b": 107,
"c": "",
"h": [
"af41"
],
"g": "dscp-fc-map"
},
{
"d": "mobility-silver-new",
"f": 4,
"e": "l1",
"b": 107,
"c": "",
"h": [
"af11",
"af21",
"af31"
],
"g": "dscp-fc-map"
}
]
}
答案 1 :(得分:1)
您的代码无法正常工作的原因是,当您遍历各行并将其插入目标字典res
时,还需要将value
设置为子字典,并且在此子词典中,行将始终位于相同的键“ b”,“ c”,“ d”等下。换句话说,底部的行将覆盖顶部的行,而您只需要Nodeb_IN_New
的第三行。
另一种方法是将value
设置为列表,然后遍历各行,如果它们属于同一qos
,则将它们追加到此列表中,这是代码:
# Original excel sheet in a list
raw_list = [
['Nodeb_IN_New', 107, 'class-default', 'mobility-platinum', 'h1', 7, 'dscp-fc-map', ['ef']],
['', 107, '', 'mobility-gold-new', 'h2', 5, 'dscp-fc-map', ['af41']],
['', 107, '', 'mobility-silver-new', 'l1', 4, 'dscp-fc-map', ['af11', 'af21', 'af31']],
['Nokia_SRAN_S1-MME_X2_IN', 102, '', 'Nokia_SRAN_mobility_platinum', 'h1', 7, 'dscp-fc-map', ['ef', 'nc1']]
]
result_dict = {}
# title, will be used as key in the result_dict
title = ""
# Loop through the input list
for row in raw_list:
# If the first value is not empty, replace the title with this new value
if row[0] != "":
# update title
title = row[0]
# Insert into the dictionary, and give an empty list as a place holder,
# later we can append to this list
result_dict[title] = []
# At this stage we can append the rest of the input row to the dictionary value
result_dict[title].append(row[1:])
print("Result dict: ")
for (key, value) in result_dict.items():
print("Key: {}".format(key))
for row in value:
print(" {}".format(row))
这是上面代码的输出:
Key: Nodeb_IN_New
[107, 'class-default', 'mobility-platinum', 'h1', 7, 'dscp-fc-map', ['ef']]
[107, '', 'mobility-gold-new', 'h2', 5, 'dscp-fc-map', ['af41']]
[107, '', 'mobility-silver-new', 'l1', 4, 'dscp-fc-map', ['af11', 'af21', 'af31']]
Key: Nokia_SRAN_S1-MME_X2_IN
[102, '', 'Nokia_SRAN_mobility_platinum', 'h1', 7, 'dscp-fc-map', ['ef', 'nc1']]