python - 将dict的一部分与重复项合并
问题描述
我是 Python 的初学者,所以请多多包涵。
我有一个看起来像这样的字典列表:
list = [
{
"name": "Bus 60",
"direction": "City",
"timeLeft": "1",
"timeNext": ""
},
{
"name": "Bus 60",
"direction": "City",
"timeLeft": "3",
"timeNext": ""
},
{
"name": "Bus 1",
"direction": "Some Place",
"timeLeft": "15",
"timeNext": ""
},
{
"name": "Bus 1",
"direction": "Some Place",
"timeLeft": "30",
"timeNext": ""
},
{
"name": "Bus 1",
"direction": "That other place",
"timeLeft": "5",
"timeNext": ""
},
]
我希望根据“名称”和“方向”合并这两者,如下所示:
new_list = [
{
"name": "Bus 60",
"direction": "City",
"timeLeft": "1",
"timeNext": "3"
},
{
"name": "Bus 1",
"direction": "Some Place",
"timeLeft": "15",
"timeNext": "30"
},
{
"name": "Bus 1",
"direction": "That other place",
"timeLeft": "5",
"timeNext": ""
},
]
我怎样才能做到这一点并了解它的实际工作原理?我已经尝试了很多带有循环的解决方案,但最终都会出现很多重复或错误的合并。
编辑:每个名称和方向的重复项永远不会超过一个。
编辑2:这是我的完整方法:
@APP.route('/api/vasttrafik/departures', methods=['POST'])
def get_departures():
""" Departures """
APP.logger.info('get_departures():')
data = request.get_json()
id_number = data['id']
current_date = date.today().strftime('%Y-%m-%d')
current_time = datetime.now().strftime('%H:%M')
# time_span = data['']
access_token = request.headers['access_token']
url = 'https://api.vasttrafik.se/bin/rest.exe/v2/departureBoard?id='\
+ id_number + '&date=' + current_date + '&time=' + current_time +\
'&format=json&timeSpan=90&maxDeparturesPerLine=2&needJourneyDetail=0'
headers = {'Authorization': 'Bearer ' + access_token}
req = requests.get(url, headers=headers)
json = req.json()
departure_board = json['DepartureBoard']
if 'error' in departure_board:
raise NotFoundException('Did not find anything')
departures = departure_board['Departure']
def departures_model(item):
def get_key_value(key):
return item[key] if key in item else ''
is_live = 'rtTime' in item
if is_live:
current_time = get_key_value('rtTime')
current_date = get_key_value('rtDate')
else:
current_time = get_key_value('time')
current_date = get_key_value('date')
direction = get_key_value('direction')
via = ''
if 'via' in direction:
direction, via = direction.split('via')
time_departure = datetime.strptime(current_date + ' ' + current_time, '%Y-%m-%d %H:%M')
time_now = datetime.now()
diff = time_departure - time_now
if time_now >= time_departure:
minutes_left = 0
else:
minutes_left = math.floor(((diff).seconds) / 60)
clock_left = item['rtTime'] if is_live else item['time']
return dict({
'accessibility': get_key_value('accessibility'),
'bgColor': get_key_value('bgColor'),
'clockLeft': clock_left,
'clockNext': '',
'timeLeft': int(minutes_left),
'timeNext': '',
'direction': direction.strip(),
'via': 'via ' + via.strip() if via != '' else via,
'name': get_key_value('name'),
'sname': get_key_value('sname'),
'type': get_key_value('type'),
'time': get_key_value('time'),
'date': get_key_value('date'),
'journeyid': get_key_value('journeyid'),
'track': get_key_value('track'),
'fgColor': get_key_value('fgColor'),
'isLive': is_live,
'night': 'night' in item,
})
mapped_departures = list(map(departures_model, departures))
def key(bus):
return bus["name"], bus["direction"]
def merge_busses(ls):
for (name, direction), busses in groupby(ls, key):
busses = list(busses)
times = [bus["timeLeft"] for bus in busses]
yield {
**busses[0],
"timeLeft": min(times, key=int),
"timeNext": max(times, key=int),
}
merge_departures = list(merge_busses(mapped_departures))
return jsonify({
'departures': merge_departures,
})
编辑 3:我刚刚发现为什么 L3viathan 和 Patrick Artner 的解决方案不起作用。它们仅在预先对总线列表进行排序时才起作用。所以我猜想 groupby 需要 dicts 是相邻的。
解决方案
这是我的解决方案:我们使用 将总线按名称-方向组合进行分组itertools.groupby
,然后生成字典,其中timeLeft
是这些总线中的最小分钟数,并且timeNext
是空字符串(如果我们只看到一个总线)或最大数量的那些公共汽车内的分钟。
from itertools import groupby
def key(bus):
return bus["name"], bus["direction"]
def merge_busses(ls):
for (name, direction), busses in groupby(sorted(ls, key=key), key):
busses = list(busses)
times = [bus["timeLeft"] for bus in busses]
yield {
**busses[0],
"timeLeft": min(times, key=int),
"timeNext": "" if len(times) == 1 else max(times, key=int),
}
像这样使用它:
new_list = list(merge_busses(mylist))
在您的示例中使用,这会产生:
[
{
"name": "Bus 60",
"direction": "City",
"timeLeft": "1",
"timeNext": "3"
},
{
"name": "Bus 1",
"direction": "Some Place",
"timeLeft": "15",
"timeNext": "30"
},
{
"name": "Bus 1",
"direction": "That other place",
"timeLeft": "5",
"timeNext": ""
}
]
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