input={11: {'perc': 0, 'name': u'B test', 'cid': 11, 'total': 0, 'pending': 0, 'complete': 0}, 10: {'perc': 0, 'name': u'C test', 'cid': 10, 'total': 0, 'pending': 0,'complete': 0}, 3: {'perc': 9, 'name': u'Atest Pre-requisites', 'cid': 3, 'total': 11, 'pending': 10, 'complete': 1}}
I want to sort this dict based on name field. I'm new in python, anyone please help me.
First, you should avoid using reserved words (such as input) as variables (now input is redefined and no longer calls the function input()).
Also, a dictionary cannot be sorted. If you don't need the keys, you can transform the dictionary into a list, and then sort it. The code would be like this:
input_dict = {11: {'perc': 0, 'name': u'B test', 'cid': 11, 'total': 0, 'pending': 0, 'complete': 0}, 10: {'perc': 0, 'name': u'C test', 'cid': 10, 'total': 0, 'pending': 0,'complete': 0}, 3: {'perc': 9, 'name': u'Atest Pre-requisites', 'cid': 3, 'total': 11, 'pending': 10, 'complete': 1}}
input_list = sorted(input_dict.values(), key=lambda x: x['name'])
print(input_list)
# prints [{'perc': 9, 'complete': 1, 'cid': 3, 'total': 11, 'pending': 10, 'name': u'Atest Pre-requisites'}, {'perc': 0, 'complete': 0, 'cid': 11, 'total': 0, 'pending': 0, 'name': u'B test'}, {'perc': 0, 'complete': 0, 'cid': 10, 'total': 0, 'pending': 0, 'name': u'C test'}]
EDIT
If you wish to keep the keys and use iteritems() as you said in the comments, use this code instead:
input_dict = {11: {'perc': 0, 'name': u'B test', 'cid': 11, 'total': 0, 'pending': 0, 'complete': 0}, 10: {'perc': 0, 'name': u'C test', 'cid': 10, 'total': 0, 'pending': 0,'complete': 0}, 3: {'perc': 9, 'name': u'Atest Pre-requisites', 'cid': 3, 'total': 11, 'pending': 10, 'complete': 1}}
input_list = sorted(input_dict.iteritems(), key=lambda x: x[1]['name'])
print(input_list)
# prints [(3, {'perc': 9, 'complete': 1, 'cid': 3, 'total': 11, 'pending': 10, 'name': u'Atest Pre-requisites'}), (11, {'perc': 0, 'complete': 0, 'cid': 11, 'total': 0, 'pending': 0, 'name': u'B test'}), (10, {'perc': 0, 'complete': 0, 'cid': 10, 'total': 0, 'pending': 0, 'name': u'C test'})]
Related
main_dict = {
'NSE:ACC': {'average_price': 0,
'buy_quantity': 0,
'depth': {'buy': [{'orders': 0, 'price': 0, 'quantity': 0},
{'orders': 0, 'price': 0, 'quantity': 0},
{'orders': 0, 'price': 0, 'quantity': 0},
{'orders': 0, 'price': 0, 'quantity': 0},
{'orders': 0, 'price': 0, 'quantity': 0}],
'sell': [{'orders': 0, 'price': 0, 'quantity': 0},
{'orders': 0, 'price': 0, 'quantity': 0},
{'orders': 0, 'price': 0, 'quantity': 0},
{'orders': 0, 'price': 0, 'quantity': 0},
{'orders': 0, 'price': 0, 'quantity': 0}]},
'instrument_token': 5633,
'last_price': 2488.9,
'last_quantity': 0,
'last_trade_time': '2022-09-23 15:59:10',
'lower_circuit_limit': 2240.05,
'net_change': 0,
'ohlc': {'close': 2555.7,
'high': 2585.5,
'low': 2472.2,
'open': 2575},
'oi': 0,
'oi_day_high': 0,
'oi_day_low': 0,
'sell_quantity': 0,
'timestamp': '2022-09-23 18:55:17',
'upper_circuit_limit': 2737.75,
'volume': 0},
}
convert dict to pandas dataframe
for example:
symbol last_price net_change Open High Low Close
NSE:ACC 2488.9 0 2575 2585.5 2472.2 2555.7
I am trying pd.DataFrame.from_dict(main_dict)
but it does not work.
please give the best suggestion.
I would first select the necessary data from your dict and then pass that as input to pd.DataFrame()
df_input = [{
"symbol": symbol,
"last_price": main_dict.get(symbol).get("last_price"),
"net_change": main_dict.get(symbol).get("net_change"),
"open": main_dict.get(symbol).get("ohlc").get("open"),
"high": main_dict.get(symbol).get("ohlc").get("high"),
"low": main_dict.get(symbol).get("ohlc").get("low"),
"close": main_dict.get(symbol).get("ohlc").get("close")
} for symbol in main_dict]
import pandas as pd
df = pd.DataFrame(df_input)
I have a list with barline ticks and midi notes that can overlap the barlines. So I made a list of 'barlineticks':
barlinepos = [0, 768.0, 1536.0, 2304.0, 3072.0, 3840.0, 4608.0, 5376.0, 6144.0, 6912.0, 0, 576.0, 1152.0, 1728.0, 2304.0, 2880.0, 3456.0, 4032.0, 4608.0, 5184.0, 5760.0, 6336.0, 6912.0, 7488.0]
And a MidiFile:
{'type': 'time_signature', 'numerator': 4, 'denominator': 4, 'time': 0, 'duration': 768, 'ID': 0}
{'type': 'set_tempo', 'tempo': 500000, 'time': 0, 'ID': 1}
{'type': 'track_name', 'name': 'Tempo Track', 'time': 0, 'ID': 2}
{'type': 'track_name', 'name': 'New Instrument', 'time': 0, 'ID': 3}
{'type': 'note_on', 'time': 0, 'channel': 0, 'note': 48, 'velocity': 100, 'ID': 4, 'duration': 956}
{'type': 'time_signature', 'numerator': 3, 'denominator': 4, 'time': 768, 'duration': 6911, 'ID': 5}
{'type': 'note_on', 'time': 768, 'channel': 0, 'note': 46, 'velocity': 100, 'ID': 6, 'duration': 575}
{'type': 'note_off', 'time': 956, 'channel': 0, 'note': 48, 'velocity': 0, 'ID': 7}
{'type': 'note_off', 'time': 1343, 'channel': 0, 'note': 46, 'velocity': 0, 'ID': 8}
{'type': 'end_of_track', 'time': 7679, 'ID': 9}
And I want to check if the midi note is overlapping a barline. Every note_on message has a 'time' and a 'duration' value. I have to check if one of the barlineticks(in the list) is inside the range of the note('time' and 'duration'). I tried:
if barlinepos in range(0, 956):
print(True)
Of course this doesn't work because barlinepos is a list. How can I check if one of the values in the list results in True?
Simple iteration to solve the requirement:
for i in midifile:
start, end = i["time"], i["time"]+i["duration"]
for j in barlinepos:
if j >= start and j<= end:
print(True)
break
print(False)
I have two lists (with dicts in it):
old_device_data_list = [{'_id': ObjectId('5f48c8e34545fac49fbff5'), 'device_id': 5, 'time': datetime.datetime(2020, 8, 26, 9, 5, 39, 827000), 'values': {'count': 100, 'late': 0, 'max': 0, 'min': 0, 'on_time': 100, 'sum': 100}}]
result = [{'_id': ObjectId('5f48c8e3997640fac49fbff5'), 'device_id': 5, 'time': datetime.datetime(2020, 8, 26, 9, 5, 39, 827000), 'values': {'count': 100, 'late': 0, 'max': 0, 'min': 0, 'on_time': 100, 'sum': 100}}, {'_id': ObjectId('5f48c8e3997640fac49fbff6'), 'device_id': 4, 'time': datetime.datetime(2020, 8, 26, 9, 5, 39, 827000), 'values': {'count': 180, 'late': 0, 'max': 0, 'min': 0, 'on_time': 180, 'sum': 180}}, {'_id': ObjectId('5f48c8e3997640fac49fbff8'), 'device_id': 3, 'time': datetime.datetime(2020, 8, 27, 9, 5, 39, 827000), 'values': {'count': 50, 'late': 0, 'max': 0, 'min': 0, 'on_time': 50, 'sum': 50}}, {'_id': ObjectId('5f48c8e3997640fac49fbff7'), 'device_id': 4, 'time': datetime.datetime(2020, 8, 27, 9, 5, 39, 827000), 'values': {'count': 120, 'late': 0, 'max': 0, 'min': 0, 'on_time': 120, 'sum': 120}}, {'_id': ObjectId('5f48c8e3997640fac49fbff9'), 'device_id': 3, 'time': datetime.datetime(2020, 8, 28, 9, 5, 39, 827000), 'values': {'count': 210, 'late': 0, 'max': 0, 'min': 0, 'on_time': 210, 'sum': 210}}]
I want to delete the dicts from the old_device_data_list out of the result list. I tried it with numpy with:
numpy.setdiff1d(result, old_device_data_list)
Then I got error:
TypeError: '<' not supported between instances of 'dict' and 'dict'
The description of numpy.setdiff1d says:
Return the sorted, unique values in ar1 that are not in ar2.
In order to sort the values, it needs to compare them using the < operator. But dictionaries cannot be compared like this. The relation "smaller than" is not defined for dictionaries.
NumPy is designed for working with numeric values, not for arbitrary Python data structures.
You could use a simple list comprehension to create a list of those dictionaries that are in result but not in old_device_data_list:
result = [d for d in result if d not in old_device_data_list]
I have all_data dataframe. I want to replace some categorical values in certain columns with numerical values. I'm trying to use this nested dictionary notation (I've checked that the brackets and curly brackets are in place, I don't think that's the issue):
all_data = all_data.replace({'Street': {'Pave': 1, 'Grvl': 0}},
{'LotShape': {'IR3': 1, 'IR2': 2, 'IR1': 3, 'Reg': 4}},
{'Utilities': {'ELO': 0, 'NoSeWa': 0, 'NoSewr': 0, 'AllPub': 1}},
{'LandSlope': {'Sev': 1, 'Mod': 2, 'Gtl': 3}},
{'ExterQual': {'Po': 1, 'Fa': 2, 'TA': 3, 'Gd': 4, 'Ex': 5}},
{'ExterCond': {'Po': 1, 'Fa': 2, 'TA': 3, 'Gd': 4, 'Ex': 5}},
{'BsmtQual': {'NA': 0, 'Po': 1, 'Fa': 2, 'TA': 3, 'Gd': 4,'Ex': 5}},
{'BsmtCond': {'NA': 0, 'Po': 1, 'Fa': 2, 'TA': 3, 'Gd': 4,'Ex': 5}},
{'BsmtExposure': {'NA': 0, 'No': 1, 'Mn': 2, 'Av': 3, 'Gd': 4}},
{'BsmtFinType1': {'NA': 0, 'Unf': 1, 'LwQ': 2, 'Rec': 3, 'BLQ': 4, 'ALQ': 5, 'GLQ': 6}},
{'BsmtFinType2': {'NA': 0, 'Unf': 1,'LwQ': 2,'Rec': 3, 'BLQ': 4,'ALQ': 5, 'GLQ': 6}},
{'HeatingQC': {'Po': 1,'Fa': 2,'TA': 3,'Gd': 4,'Ex': 5}},
{'CentralAir': {'No': 0,'Yes': 1}},
{'KitchenQual': {'Po': 1,'Fa': 2,'TA': 3,'Gd': 4,'Ex': 5}},
{'Functional': {'Sal': -7,'Sev': -6,'Maj1': -5,'Maj2': -4,'Mod': -3,'Min2': -2,'Min1': -1,
'Typ': 0}},
{'FireplaceQu': {'NA': 0,'Po': 1,'Fa': 2,'TA': 3,'Gd': 4,'Ex': 5}},
{'GarageFinish': {'NA': 0,'Unf': 1,'RFn': 2, 'Fin': 3}},
{'GarageQual': {'NA': 0, 'Po': 1,'Fa': 2, 'TA': 3,'Gd': 4, 'Ex': 5}},
{'GarageCond': {'NA': 0,'Po': 1,'Fa': 2,'TA': 3,'Gd': 4,'Ex': 5}},
{'PavedDrive': {'N': 0,'P': 0, 'Y': 1}},
{'Fence': {'NA': 0, 'MnWw': 1,'GdWo': 2,'MnPrv': 3,'GdPrv': 4}},
{'SaleCondition': {'Abnorml': 1, 'Alloca': 1, 'AdjLand': 1, 'Family': 1, 'Normal': 0,
'Partial': 0}}
)
Error:
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-40-f9c9c28b7237> in <module>()
22 {'Fence': {'NA': 0, 'MnWw': 1,'GdWo': 2,'MnPrv': 3,'GdPrv': 4}},
23 {'SaleCondition': {'Abnorml': 1, 'Alloca': 1, 'AdjLand': 1, 'Family': 1, 'Normal': 0,
---> 24 'Partial': 0}}
25 )
TypeError: replace() takes from 1 to 8 positional arguments but 23 were given
If I remove the 'SaleCondition' row from the above code, the error is again there but this time referring to 'Fence', and so on, for each line of code from bottom up. I've googled but have no idea what this means. Help MUCH appreciated.
You should do something like :
df.replace({'Fence':{'NA': 0, 'MnWw': 1,'GdWo': 2,'MnPrv': 3,'GdPrv': 4},'SaleCondition':{'Abnorml': 1, 'Alloca': 1, 'AdjLand': 1, 'Family': 1, 'Normal': 0,
'Partial': 0}})
the format should be .replace({'col1':{},'col2':{}}) not .replace({'col1':{}},{'col2':{}})
I have this :
[
[{ 'position': 1, 'user_id': 2, 'value': 4, 'points': 100}],
[{ 'position': 2, 'user_id': 6, 'value': 3, 'points': 88}],
[{ 'position': 3, 'user_id': 5, 'value': 2, 'points': 77}],
[{ 'position': 4, 'user_id': 7, 'value': 1, 'points': 66}],
[{ 'position': 5, 'user_id': 3, 'value': 1, 'points': 9}],
[{ 'position': 6, 'user_id': 11, 'value': 0, 'points': 9}],
[{ 'position': 7, 'user_id': 1, 'value': 0, 'points': 3}],
[{ 'position': 8, 'user_id': 10, 'value': 0, 'points': 3}],
[{ 'position': 9, 'user_id': 4, 'value': 0, 'points': 2}],
[{ 'position': 10, 'user_id': 8, 'value': 0, 'points': 2}]
]
is organized by points.
The idea is to choose the user_id and generate a new list with the selected 5 users.
Example:
user_id=3:
[{ 'position': 3, 'user_id': 5, 'value': 2, 'points': 77}],
[{ 'position': 4, 'user_id': 7, 'value': 1, 'points': 66}],
[{ 'position': 5, 'user_id': 3, 'value': 1, 'points': 9}],
[{ 'position': 6, 'user_id': 11, 'value': 0, 'points': 9}],
[{ 'position': 7, 'user_id': 1, 'value': 0, 'points': 3}]
It returns user_id 3 in the middle with 2 users hight and 2 users lower
user_id=2
[{ 'position': 1, 'user_id': 2, 'value': 4, 'points': 100}],
[{ 'position': 2, 'user_id': 6, 'value': 3, 'points': 88}],
[{ 'position': 3, 'user_id': 5, 'value': 2, 'points': 77}],
[{ 'position': 4, 'user_id': 7, 'value': 1, 'points': 66}],
[{ 'position': 5, 'user_id': 3, 'value': 1, 'points': 9}],
As user_id hasn't higher users it returns 4 lower users. So is always same logic.
user_id=9:
[{ 'position': 6, 'user_id': 11, 'value': 0, 'points': 9}],
[{ 'position': 7, 'user_id': 1, 'value': 0, 'points': 3}],
[{ 'position': 8, 'user_id': 10, 'value': 0, 'points': 3}],
[{ 'position': 9, 'user_id': 4, 'value': 0, 'points': 2}],
[{ 'position': 10, 'user_id': 8, 'value': 0, 'points': 2}]
on user_id=9 We only have 1 user lower so we add 3 higher users
If for example we just have 2 users in list, it should return that 2 users.
Main rules:
If we have 5 users or more, as to return 5 users.
if we have 4 users, as to return 4 users
How is a good way to do it?
thanks
This is basically only an update of my answer to your original question.
a = [
[{ 'position': 1, 'user_id': 2, 'value': 4, 'points': 100}],
[{ 'position': 2, 'user_id': 6, 'value': 3, 'points': 88}],
[{ 'position': 3, 'user_id': 5, 'value': 2, 'points': 77}],
[{ 'position': 4, 'user_id': 7, 'value': 1, 'points': 66}],
[{ 'position': 5, 'user_id': 3, 'value': 1, 'points': 9}],
[{ 'position': 6, 'user_id': 11, 'value': 0, 'points': 9}],
[{ 'position': 7, 'user_id': 1, 'value': 0, 'points': 3}],
[{ 'position': 8, 'user_id': 10, 'value': 0, 'points': 3}],
[{ 'position': 9, 'user_id': 4, 'value': 0, 'points': 2}],
[{ 'position': 10, 'user_id': 8, 'value': 0, 'points': 2}]
]
# Sort it if not already sorted
# a.sort(key=lambda x: x[0]['position'])
def find_index(l, user_id):
i = 0
while l[i][0]['user_id'] != user_id:
i += 1
return i
def get_subset(l, i):
return l[:(i + 1 + max(2, 4 - i))][-5:]
get_subset(a, find_index(a, 3))