Convert `DataFrame.groupby()` into dictionary (and then reverse it) - python

Say I have the following DataFrame() where I have repeated observations per individual (column id_ind). Hence, first two rows belong the first individual, the third and fourth rows belong to the second individual, and so forth...
import pandas as pd
X = pd.DataFrame.from_dict({'x1_1': {0: -0.1766214634108258, 1: 1.645852185286492, 2: -0.13348860101031038, 3: 1.9681043689968933, 4: -1.7004428240831382, 5: 1.4580091413853749, 6: 0.06504113741068565, 7: -1.2168493676768384, 8: -0.3071304478616376, 9: 0.07121332925591593}, 'x1_2': {0: -2.4207773498298844, 1: -1.0828751040719462, 2: 2.73533787008624, 3: 1.5979611987152071, 4: 0.08835542172064115, 5: 1.2209786277076156, 6: -0.44205979195950784, 7: -0.692872860268244, 8: 0.0375521181289943, 9: 0.4656030062266639}, 'x1_3': {0: -1.548320898226322, 1: 0.8457342014424675, 2: -0.21250514722879738, 3: 0.5292389938329516, 4: -2.593946520223666, 5: -0.6188958526077123, 6: 1.6949245117526974, 7: -1.0271341091035742, 8: 0.637561891142571, 9: -0.7717170035055559}, 'x2_1': {0: 0.3797245517345564, 1: -2.2364391598508835, 2: 0.6205947900678905, 3: 0.6623865847688559, 4: 1.562036259999875, 5: -0.13081282910947759, 6: 0.03914373833251773, 7: -0.995761652421108, 8: 1.0649494418154162, 9: 1.3744782478849122}, 'x2_2': {0: -0.5052556836786106, 1: 1.1464291788297152, 2: -0.5662380273138174, 3: 0.6875729143723538, 4: 0.04653136473130827, 5: -0.012885303852347407, 6: 1.5893672346098884, 7: 0.5464286050059511, 8: -0.10430829457707284, 9: -0.5441755265313813}, 'x2_3': {0: -0.9762973303149007, 1: -0.983731467806563, 2: 1.465827578266328, 3: 0.5325950414202745, 4: -1.4452121324204903, 5: 0.8148816373643869, 6: 0.470791989780882, 7: -0.17951636294180473, 8: 0.7351814781280054, 9: -0.28776723200679066}, 'x3_1': {0: 0.12751822396637064, 1: -0.21926633684030983, 2: 0.15758799357206943, 3: 0.5885412224632464, 4: 0.11916562911189271, 5: -1.6436210334529249, 6: -0.12444368631987467, 7: 1.4618564171802453, 8: 0.6847234328916137, 9: -0.23177118858569187}, 'x3_2': {0: -0.6452955690715819, 1: 1.052094761527654, 2: 0.20190339195326157, 3: 0.6839430295237913, 4: -0.2607691613858866, 5: 0.3315513026670213, 6: 0.015901139336566113, 7: 0.15243420084881903, 8: -0.7604225072161022, 9: -0.4387652927008854}, 'x3_3': {0: -1.067058994377549, 1: 0.8026914180717286, 2: -1.9868531745912268, 3: -0.5057770735303253, 4: -1.6589569342151713, 5: 0.358172252880764, 6: 1.9238983803281329, 7: 2.2518318810978246, 8: -1.2781475121874357, 9: -0.7103081175166167}})
Y = pd.DataFrame.from_dict({'CHOICE': {0: 1.0, 1: 1.0, 2: 2.0, 3: 2.0, 4: 3.0, 5: 2.0, 6: 1.0, 7: 1.0, 8: 2.0, 9: 2.0}})
Z = pd.DataFrame.from_dict({'z1': {0: 2.4196730570917233, 1: 2.4196730570917233, 2: 2.822802255159467, 3: 2.822802255159467, 4: 2.073171091633643, 5: 2.073171091633643, 6: 2.044165101485163, 7: 2.044165101485163, 8: 2.4001241292606275, 9: 2.4001241292606275}, 'z2': {0: 0.0, 1: 0.0, 2: 0.0, 3: 0.0, 4: 1.0, 5: 1.0, 6: 1.0, 7: 1.0, 8: 0.0, 9: 0.0}, 'z3': {0: 1.0, 1: 1.0, 2: 1.0, 3: 1.0, 4: 2.0, 5: 2.0, 6: 2.0, 7: 2.0, 8: 3.0, 9: 3.0}})
id = pd.DataFrame.from_dict({'id_choice': {0: 1.0, 1: 2.0, 2: 3.0, 3: 4.0, 4: 5.0, 5: 6.0, 6: 7.0, 7: 8.0, 8: 9.0, 9: 10.0}, 'id_ind': {0: 1.0, 1: 1.0, 2: 2.0, 3: 2.0, 4: 3.0, 5: 3.0, 6: 4.0, 7: 4.0, 8: 5.0, 9: 5.0}} )
# Create a dataframe with all the data
data = pd.concat([id, X, Z, Y], axis=1)
print(data.head(4))
# id_choice id_ind x1_1 x1_2 x1_3 x2_1 x2_2 \
# 0 1.0 1.0 -0.176621 -2.420777 -1.548321 0.379725 -0.505256
# 1 2.0 1.0 1.645852 -1.082875 0.845734 -2.236439 1.146429
# 2 3.0 2.0 -0.133489 2.735338 -0.212505 0.620595 -0.566238
# 3 4.0 2.0 1.968104 1.597961 0.529239 0.662387 0.687573
#
# x2_3 x3_1 x3_2 x3_3 z1 z2 z3 CHOICE
# 0 -0.976297 0.127518 -0.645296 -1.067059 2.419673 0.0 1.0 1.0
# 1 -0.983731 -0.219266 1.052095 0.802691 2.419673 0.0 1.0 1.0
# 2 1.465828 0.157588 0.201903 -1.986853 2.822802 0.0 1.0 2.0
# 3 0.532595 0.588541 0.683943 -0.505777 2.822802 0.0 1.0 2.0
I want to perform two operations.
First, I want to convert the DataFrame data into a dictionary of DataFrame()s where the keys are the number of individuals (in this particular case, numbers ranging from 1.0 to 5.0.). I've done this below as suggested here. Unfortunately, I am getting a dictionary of numpy values and not a dictionary of DataFrame()s.
# Create a dictionary with the data for each individual
data_dict = data.set_index('id_ind').groupby('id_ind').apply(lambda x : x.to_numpy().tolist()).to_dict()
print(data_dict.keys())
# dict_keys([1.0, 2.0, 3.0, 4.0, 5.0])
print(data_dict[1.0])
#[[1.0, -0.1766214634108258, -2.4207773498298844, -1.548320898226322, 0.3797245517345564, -0.5052556836786106, -0.9762973303149007, 0.12751822396637064, -0.6452955690715819, -1.067058994377549, 2.4196730570917233, 0.0, 1.0, 1.0], [2.0, 1.645852185286492, -1.0828751040719462, 0.8457342014424675, -2.2364391598508835, 1.1464291788297152, -0.983731467806563, -0.21926633684030983, 1.052094761527654, 0.8026914180717286, 2.4196730570917233, 0.0, 1.0, 1.0]]
Second, I want to recover the original DataFrame data reversing the previous operation. The naive approach is as follows. However, it is, of course, not producing the expected result.
# Naive approach
res = pd.DataFrame.from_dict(data_dict, orient='index')
print(res)
# 0 1
#1.0 [1.0, -0.1766214634108258, -2.4207773498298844... [2.0, 1.645852185286492, -1.0828751040719462, ...
#2.0 [3.0, -0.13348860101031038, 2.73533787008624, ... [4.0, 1.9681043689968933, 1.5979611987152071, ...
#3.0 [5.0, -1.7004428240831382, 0.08835542172064115... [6.0, 1.4580091413853749, 1.2209786277076156, ...
#4.0 [7.0, 0.06504113741068565, -0.4420597919595078... [8.0, -1.2168493676768384, -0.692872860268244,...
#5.0 [9.0, -0.3071304478616376, 0.0375521181289943,... [10.0, 0.07121332925591593, 0.4656030062266639...


This solution was inspired by #mozway comments.
# Create a dictionary with the data for each individual
data_dict = dict(list(data.groupby('id_ind')))
# Convert the dictionary into a dataframe
res = pd.concat(data_dict, axis=0).reset_index(drop=True)
print(res.head(4))
# id_choice id_ind x1_1 x1_2 x1_3 x2_1 x2_2 \
#0 1.0 1.0 -0.176621 -2.420777 -1.548321 0.379725 -0.505256
#1 2.0 1.0 1.645852 -1.082875 0.845734 -2.236439 1.146429
#2 3.0 2.0 -0.133489 2.735338 -0.212505 0.620595 -0.566238
#3 4.0 2.0 1.968104 1.597961 0.529239 0.662387 0.687573
#
# x2_3 x3_1 x3_2 x3_3 z1 z2 z3 CHOICE
#0 -0.976297 0.127518 -0.645296 -1.067059 2.419673 0.0 1.0 1.0
#1 -0.983731 -0.219266 1.052095 0.802691 2.419673 0.0 1.0 1.0
#2 1.465828 0.157588 0.201903 -1.986853 2.822802 0.0 1.0 2.0
#3 0.532595 0.588541 0.683943 -0.505777 2.822802 0.0 1.0 2.0

Related

Flatting out a multiindex dataframe

I have a df:
df = pd.DataFrame.from_dict({('group', ''): {0: 'A',
1: 'A',
2: 'A',
3: 'A',
4: 'A',
5: 'A',
6: 'A',
7: 'A',
8: 'A',
9: 'B',
10: 'B',
11: 'B',
12: 'B',
13: 'B',
14: 'B',
15: 'B',
16: 'B',
17: 'B',
18: 'all',
19: 'all'},
('category', ''): {0: 'Amazon',
1: 'Apple',
2: 'Facebook',
3: 'Google',
4: 'Netflix',
5: 'Tesla',
6: 'Total',
7: 'Uber',
8: 'total',
9: 'Amazon',
10: 'Apple',
11: 'Facebook',
12: 'Google',
13: 'Netflix',
14: 'Tesla',
15: 'Total',
16: 'Uber',
17: 'total',
18: 'Total',
19: 'total'},
(pd.Timestamp('2020-06-29'), 'last_sales'): {0: 195.0,
1: 61.0,
2: 106.0,
3: 61.0,
4: 37.0,
5: 13.0,
6: 954.0,
7: 4.0,
8: 477.0,
9: 50.0,
10: 50.0,
11: 75.0,
12: 43.0,
13: 17.0,
14: 14.0,
15: 504.0,
16: 3.0,
17: 252.0,
18: 2916.0,
19: 2916.0},
(pd.Timestamp('2020-06-29'), 'sales'): {0: 1268.85,
1: 18274.385000000002,
2: 19722.65,
3: 55547.255,
4: 15323.800000000001,
5: 1688.6749999999997,
6: 227463.23,
7: 1906.0,
8: 113731.615,
9: 3219.6499999999996,
10: 15852.060000000001,
11: 17743.7,
12: 37795.15,
13: 5918.5,
14: 1708.75,
15: 166349.64,
16: 937.01,
17: 83174.82,
18: 787625.7400000001,
19: 787625.7400000001},
(pd.Timestamp('2020-06-29'), 'difference'): {0: 0.0,
1: 0.0,
2: 0.0,
3: 0.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 0.0,
8: 0.0,
9: 0.0,
10: 0.0,
11: 0.0,
12: 0.0,
13: 0.0,
14: 0.0,
15: 0.0,
16: 0.0,
17: 0.0,
18: 0.0,
19: 0.0},
(pd.Timestamp('2020-07-06'), 'last_sales'): {0: 26.0,
1: 39.0,
2: 79.0,
3: 49.0,
4: 10.0,
5: 10.0,
6: 436.0,
7: 5.0,
8: 218.0,
9: 89.0,
10: 34.0,
11: 133.0,
12: 66.0,
13: 21.0,
14: 20.0,
15: 732.0,
16: 3.0,
17: 366.0,
18: 2336.0,
19: 2336.0},
(pd.Timestamp('2020-07-06'), 'sales'): {0: 3978.15,
1: 12138.96,
2: 19084.175,
3: 40033.46000000001,
4: 4280.15,
5: 1495.1,
6: 165548.29,
7: 1764.15,
8: 82774.145,
9: 8314.92,
10: 12776.649999999996,
11: 28048.075,
12: 55104.21000000002,
13: 6962.844999999999,
14: 3053.2000000000003,
15: 231049.11000000002,
16: 1264.655,
17: 115524.55500000001,
18: 793194.8000000002,
19: 793194.8000000002},
(pd.Timestamp('2020-07-06'), 'difference'): {0: 0.0,
1: 0.0,
2: 0.0,
3: 0.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 0.0,
8: 0.0,
9: 0.0,
10: 0.0,
11: 0.0,
12: 0.0,
13: 0.0,
14: 0.0,
15: 0.0,
16: 0.0,
17: 0.0,
18: 0.0,
19: 0.0},
(pd.Timestamp('2021-06-28'), 'last_sales'): {0: 96.0,
1: 56.0,
2: 106.0,
3: 44.0,
4: 34.0,
5: 13.0,
6: 716.0,
7: 9.0,
8: 358.0,
9: 101.0,
10: 22.0,
11: 120.0,
12: 40.0,
13: 13.0,
14: 8.0,
15: 610.0,
16: 1.0,
17: 305.0,
18: 2652.0,
19: 2652.0},
(pd.Timestamp('2021-06-28'), 'sales'): {0: 5194.95,
1: 19102.219999999994,
2: 22796.420000000002,
3: 30853.115,
4: 11461.25,
5: 992.6,
6: 188143.41,
7: 3671.15,
8: 94071.705,
9: 6022.299999999998,
10: 7373.6,
11: 33514.0,
12: 35943.45,
13: 4749.000000000001,
14: 902.01,
15: 177707.32,
16: 349.3,
17: 88853.66,
18: 731701.46,
19: 731701.46},
(pd.Timestamp('2021-06-28'), 'difference'): {0: 0.0,
1: 0.0,
2: 0.0,
3: 0.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 0.0,
8: 0.0,
9: 0.0,
10: 0.0,
11: 0.0,
12: 0.0,
13: 0.0,
14: 0.0,
15: 0.0,
16: 0.0,
17: 0.0,
18: 0.0,
19: 0.0},
(pd.Timestamp('2021-07-07'), 'last_sales'): {0: 45.0,
1: 47.0,
2: 87.0,
3: 45.0,
4: 13.0,
5: 8.0,
6: 494.0,
7: 2.0,
8: 247.0,
9: 81.0,
10: 36.0,
11: 143.0,
12: 56.0,
13: 9.0,
14: 9.0,
15: 670.0,
16: 1.0,
17: 335.0,
18: 2328.0,
19: 2328.0},
(pd.Timestamp('2021-07-07'), 'sales'): {0: 7556.414999999998,
1: 14985.05,
2: 16790.899999999998,
3: 36202.729999999996,
4: 4024.97,
5: 1034.45,
6: 163960.32999999996,
7: 1385.65,
8: 81980.16499999998,
9: 5600.544999999999,
10: 11209.92,
11: 32832.61,
12: 42137.44500000001,
13: 3885.1499999999996,
14: 1191.5,
15: 194912.34000000003,
16: 599.0,
17: 97456.17000000001,
18: 717745.3400000001,
19: 717745.3400000001},
(pd.Timestamp('2021-07-07'), 'difference'): {0: 0.0,
1: 0.0,
2: 0.0,
3: 0.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 0.0,
8: 0.0,
9: 0.0,
10: 0.0,
11: 0.0,
12: 0.0,
13: 0.0,
14: 0.0,
15: 0.0,
16: 0.0,
17: 0.0,
18: 0.0,
19: 0.0}}).set_index(['group','category'])
I am trying to sort of flatten it so it would no longer be a multiindex df. As there are several dates I try to select one:
df.loc[:,'2020-06-29 00:00:00']
But this gives me an error :
KeyError: '2020-06-29 00:00:00'
I am trying to make it that the first week ( and my final output ) of 2020-06-29 would look like this :
group category last_sales sales difference
A Amazon 195.00 1,268.85 0.00
A Apple 61.00 18,274.39 0.00
A Facebook 106.00 19,722.65 0.00
A Google 61.00 55,547.25 0.00
A Netflix 37.00 15,323.80 0.00
A Tesla 13.00 1,688.67 0.00
A Total 954.00 227,463.23 0.00
A Uber 4.00 1,906.00 0.00
A total 477.00 113,731.62 0.00
B Amazon 0.00 3,219.65 0.00
B Apple 50.00 15,852.06 0.00
B Facebook 75.00 17,743.70 0.00
B Google 43.00 37,795.15 0.00
B Netflix 17.00 5,918.50 0.00
B Tesla 14.00 1,708.75 0.00
B Total 504.00 166,349.64 0.00
B Uber 3.00 937.01 0.00
B total 252.00 83,174.82 0.00
all Total 2,916.00 787,625.74 0.00

try via pd.to_dateime():
out=df.loc[:,pd.to_datetime('2020-06-29 00:00:00')]
#out=df.loc[:,pd.to_datetime('2020-06-29 00:00:00')].reset_index()
OR
try via pd.Timestamp()
out=df.loc[:,pd.Timestamp('2020-06-29 00:00:00')]
#out=df.loc[:,pd.Timestamp('2020-06-29 00:00:00')].reset_index()
The 0th level of your column is Timestamp and you can verify that by:
df.columns.to_numpy()
#output
array([(Timestamp('2020-06-29 00:00:00'), 'last_sales'),
(Timestamp('2020-06-29 00:00:00'), 'sales'),
(Timestamp('2020-06-29 00:00:00'), 'difference'),
(Timestamp('2020-07-06 00:00:00'), 'last_sales'),
(Timestamp('2020-07-06 00:00:00'), 'sales'),
(Timestamp('2020-07-06 00:00:00'), 'difference'),
(Timestamp('2021-06-28 00:00:00'), 'last_sales'),
(Timestamp('2021-06-28 00:00:00'), 'sales'),
(Timestamp('2021-06-28 00:00:00'), 'difference'),
(Timestamp('2021-07-07 00:00:00'), 'last_sales'),
(Timestamp('2021-07-07 00:00:00'), 'sales'),
(Timestamp('2021-07-07 00:00:00'), 'difference')], dtype=object)
output of out:
last_sales sales difference
group category
A Amazon 195.0 1268.850 0.0
Apple 61.0 18274.385 0.0
Facebook 106.0 19722.650 0.0
Google 61.0 55547.255 0.0
Netflix 37.0 15323.800 0.0
Tesla 13.0 1688.675 0.0
Total 954.0 227463.230 0.0
Uber 4.0 1906.000 0.0
total 477.0 113731.615 0.0
B Amazon 50.0 3219.650 0.0
Apple 50.0 15852.060 0.0
Facebook 75.0 17743.700 0.0
Google 43.0 37795.150 0.0
Netflix 17.0 5918.500 0.0
Tesla 14.0 1708.750 0.0
Total 504.0 166349.640 0.0
Uber 3.0 937.010 0.0
total 252.0 83174.820 0.0
all Total 2916.0 787625.740 0.0
total 2916.0 787625.740 0.0
NOTE:
There is no need of providing a tuple in .loc[] because you are selecting the 0th level

I’m also getting a KeyError, but if you use a Timestamp object to index the first-level columns, it seems to work:
>>> df[pd.Timestamp('2020-06-29 00:00:00')]
last_sales sales difference
group category
A Amazon 195.0 1268.850 0.0
Apple 61.0 18274.385 0.0
Facebook 106.0 19722.650 0.0
Google 61.0 55547.255 0.0
Netflix 37.0 15323.800 0.0
Tesla 13.0 1688.675 0.0
Total 954.0 227463.230 0.0
Uber 4.0 1906.000 0.0
total 477.0 113731.615 0.0
B Amazon 50.0 3219.650 0.0
Apple 50.0 15852.060 0.0
Facebook 75.0 17743.700 0.0
Google 43.0 37795.150 0.0
Netflix 17.0 5918.500 0.0
Tesla 14.0 1708.750 0.0
Total 504.0 166349.640 0.0
Uber 3.0 937.010 0.0
total 252.0 83174.820 0.0
all Total 2916.0 787625.740 0.0
total 2916.0 787625.740 0.0
Otherwise you could use .xs which will then also allow you more flexibility, e.g. selecting in the second level of columns and so on:
>>> df.xs(pd.Timestamp('2020-06-29 00:00:00'), axis='columns', level=0)
last_sales sales difference
group category
A Amazon 195.0 1268.850 0.0
Apple 61.0 18274.385 0.0
Facebook 106.0 19722.650 0.0
Google 61.0 55547.255 0.0
Netflix 37.0 15323.800 0.0
Tesla 13.0 1688.675 0.0
Total 954.0 227463.230 0.0
Uber 4.0 1906.000 0.0
total 477.0 113731.615 0.0
B Amazon 50.0 3219.650 0.0
Apple 50.0 15852.060 0.0
Facebook 75.0 17743.700 0.0
Google 43.0 37795.150 0.0
Netflix 17.0 5918.500 0.0
Tesla 14.0 1708.750 0.0
Total 504.0 166349.640 0.0
Uber 3.0 937.010 0.0
total 252.0 83174.820 0.0
all Total 2916.0 787625.740 0.0
total 2916.0 787625.740 0.0
You can then add .drop(index=[('all', 'total')]) to remove the second total line, and possible .reset_index()
The way to do it with .loc[] is to provide a tuple, with the first item being a Timestamp object and the second an empty slice. However this will keep the 2 levels of indexing, so it is not what you want:
>>> df.loc[:, (pd.Timestamp('2020-06-29 00:00:00'), slice(None))].head(2)
2020-06-29 00:00:00
last_sales sales difference
group category
A Amazon 195.0 1268.850 0.0
Apple 61.0 18274.385 0.0

Adding a total per level-2 index in a multiindex pandas dataframe

I have a dataframe:
df_full = pd.DataFrame.from_dict({('group', ''): {0: 'A',
1: 'A',
2: 'A',
3: 'A',
4: 'A',
5: 'A',
6: 'A',
7: 'B',
8: 'B',
9: 'B',
10: 'B',
11: 'B',
12: 'B',
13: 'B'},
('category', ''): {0: 'Books',
1: 'Candy',
2: 'Pencil',
3: 'Table',
4: 'PC',
5: 'Printer',
6: 'Lamp',
7: 'Books',
8: 'Candy',
9: 'Pencil',
10: 'Table',
11: 'PC',
12: 'Printer',
13: 'Lamp'},
(pd.Timestamp('2021-06-28 00:00:00'),
'Sales_1'): {0: 9.937449997200002, 1: 30.71300000639998, 2: 58.81199999639999, 3: 25.661999978399994, 4: 3.657999996, 5: 12.0879999972, 6: 61.16600000040001, 7: 6.319439989199998, 8: 12.333119997600003, 9: 24.0544100028, 10: 24.384659998799997, 11: 1.9992000012000002, 12: 0.324, 13: 40.69122000000001},
(pd.Timestamp('2021-06-28 00:00:00'),
'Sales_2'): {0: 21.890370397789923, 1: 28.300470581874837, 2: 53.52039700062155, 3: 52.425508769690694, 4: 6.384936971649232, 5: 6.807138946302334, 6: 52.172, 7: 5.916852561, 8: 5.810764652, 9: 12.1243325, 10: 17.88071596, 11: 0.913782413, 12: 0.869207661, 13: 20.9447844},
(pd.Timestamp('2021-06-28 00:00:00'), 'last_week_sales'): {0: np.nan,
1: np.nan,
2: np.nan,
3: np.nan,
4: np.nan,
5: np.nan,
6: np.nan,
7: np.nan,
8: np.nan,
9: np.nan,
10: np.nan,
11: np.nan,
12: np.nan,
13: np.nan},
(pd.Timestamp('2021-06-28 00:00:00'), 'total_orders'): {0: 86.0,
1: 66.0,
2: 188.0,
3: 556.0,
4: 12.0,
5: 4.0,
6: 56.0,
7: 90.0,
8: 26.0,
9: 49.0,
10: 250.0,
11: 7.0,
12: 2.0,
13: 44.0},
(pd.Timestamp('2021-06-28 00:00:00'), 'total_sales'): {0: 4390.11,
1: 24825.059999999998,
2: 48592.39999999998,
3: 60629.77,
4: 831.22,
5: 1545.71,
6: 34584.99,
7: 5641.54,
8: 6798.75,
9: 13290.13,
10: 42692.68000000001,
11: 947.65,
12: 329.0,
13: 29889.65},
(pd.Timestamp('2021-07-05 00:00:00'),
'Sales_1'): {0: 13.690399997999998, 1: 38.723000005199985, 2: 72.4443400032, 3: 36.75802000560001, 4: 5.691999996, 5: 7.206999998399999, 6: 66.55265999039996, 7: 6.4613199911999954, 8: 12.845630001599998, 9: 26.032340003999998, 10: 30.1634600016, 11: 1.0203399996, 12: 1.4089999991999997, 13: 43.67116000320002},
(pd.Timestamp('2021-07-05 00:00:00'),
'Sales_2'): {0: 22.874363860953647, 1: 29.5726042895728, 2: 55.926190956481534, 3: 54.7820864335212, 4: 6.671946105284065, 5: 7.113126469779095, 6: 54.517, 7: 6.194107518, 8: 6.083562133, 9: 12.69221484, 10: 18.71872129, 11: 0.956574175, 12: 0.910216433, 13: 21.92632044},
(pd.Timestamp('2021-07-05 00:00:00'), 'last_week_sales'): {0: 4390.11,
1: 24825.059999999998,
2: 48592.39999999998,
3: 60629.77,
4: 831.22,
5: 1545.71,
6: 34584.99,
7: 5641.54,
8: 6798.75,
9: 13290.13,
10: 42692.68000000001,
11: 947.65,
12: 329.0,
13: 29889.65},
(pd.Timestamp('2021-07-05 00:00:00'), 'total_orders'): {0: 109.0,
1: 48.0,
2: 174.0,
3: 587.0,
4: 13.0,
5: 5.0,
6: 43.0,
7: 62.0,
8: 13.0,
9: 37.0,
10: 196.0,
11: 8.0,
12: 1.0,
13: 33.0},
(pd.Timestamp('2021-07-05 00:00:00'), 'total_sales'): {0: 3453.02,
1: 17868.730000000003,
2: 44707.82999999999,
3: 60558.97999999999,
4: 1261.0,
5: 1914.6000000000001,
6: 24146.09,
7: 6201.489999999999,
8: 5513.960000000001,
9: 9645.87,
10: 25086.785,
11: 663.0,
12: 448.61,
13: 26332.7}}).set_index(['group','category'])
I am trying to get a total for each column per category. So in this df example adding 2 lines below Lamp denoting the totals of each column. Red lines indicate the desired totals placement:
What I've tried:
df_out['total'] = df_out.sum(level=1).loc[:, (slice(None), 'total_sales')]
But get:
ValueError: Wrong number of items passed 4, placement implies 1
I also checked this question but could not apply it to my self.

Let us try groupby on level=0
s = df_full.groupby(level=0).sum()
s.index = pd.MultiIndex.from_product([s.index, ['Total']])
df_out = df_full.append(s).sort_index()
print(df_out)
2021-06-28 00:00:00 2021-07-05 00:00:00
Sales_1 Sales_2 last_week_sales total_orders total_sales Sales_1 Sales_2 last_week_sales total_orders total_sales
group category
A Books 9.93745 21.890370 NaN 86.0 4390.11 13.69040 22.874364 4390.11 109.0 3453.020
Candy 30.71300 28.300471 NaN 66.0 24825.06 38.72300 29.572604 24825.06 48.0 17868.730
Lamp 61.16600 52.172000 NaN 56.0 34584.99 66.55266 54.517000 34584.99 43.0 24146.090
PC 3.65800 6.384937 NaN 12.0 831.22 5.69200 6.671946 831.22 13.0 1261.000
Pencil 58.81200 53.520397 NaN 188.0 48592.40 72.44434 55.926191 48592.40 174.0 44707.830
Printer 12.08800 6.807139 NaN 4.0 1545.71 7.20700 7.113126 1545.71 5.0 1914.600
Table 25.66200 52.425509 NaN 556.0 60629.77 36.75802 54.782086 60629.77 587.0 60558.980
Total 202.03645 221.500823 0.0 968.0 175399.26 241.06742 231.457318 175399.26 979.0 153910.250
B Books 6.31944 5.916853 NaN 90.0 5641.54 6.46132 6.194108 5641.54 62.0 6201.490
Candy 12.33312 5.810765 NaN 26.0 6798.75 12.84563 6.083562 6798.75 13.0 5513.960
Lamp 40.69122 20.944784 NaN 44.0 29889.65 43.67116 21.926320 29889.65 33.0 26332.700
PC 1.99920 0.913782 NaN 7.0 947.65 1.02034 0.956574 947.65 8.0 663.000
Pencil 24.05441 12.124332 NaN 49.0 13290.13 26.03234 12.692215 13290.13 37.0 9645.870
Printer 0.32400 0.869208 NaN 2.0 329.00 1.40900 0.910216 329.00 1.0 448.610
Table 24.38466 17.880716 NaN 250.0 42692.68 30.16346 18.718721 42692.68 196.0 25086.785
Total 110.10605 64.460440 0.0 468.0 99589.40 121.60325 67.481717 99589.40 350.0 73892.415

Bokeh Hovertool stacked barchart

I have constructed a Bokeh stacked barchart by the code below. The chart shows the different tree types for the districts of Copenhagen. At the moment I have a hoverTool which shows the excat amount of trees (corrosponding to the columns with the tree names) for the tree type but I also want it to show the percentage (the columns with _pat the end), but how can I do this with the stacked bar chart?
A reduced part of the data frame:
temp=pd.DataFrame( {'bydelsnavn': {0: 'Amager Vest', 1: 'Amager Øst', 2: 'Bispebjerg', 3: 'Brønshøj-Husum', 4: 'Indre By', 5: 'Nørrebro', 6: 'Valby', 7: 'Vanløse', 8: 'Vesterbro', 9: 'Østerbro'}, 'Alder': {0: 53.0, 1: 21.0, 2: 1.0, 3: 9.0, 4: 4.0, 5: 2.0, 6: 3.0, 7: 44.0, 8: 46.0, 9: 59.0}, 'Alderm': {0: 63.0, 1: 32.0, 2: 49.0, 3: 13.0, 4: 45.0, 5: 55.0, 6: 104.0, 7: 0.0, 8: 50.0, 9: 4.0}, 'Apple': {0: 94.0, 1: 109.0, 2: 115.0, 3: 12.0, 4: 22.0, 5: 81.0, 6: 41.0, 7: 3.0, 8: 132.0, 9: 51.0}, 'Alder_p': {0: 21.9, 1: 8.68, 2: 0.41, 3: 3.72, 4: 1.65, 5: 0.83, 6: 1.24, 7: 18.18, 8: 19.01, 9: 24.38}, 'Alderm_p': {0: 15.18, 1: 7.71, 2: 11.81, 3: 3.13, 4: 10.84, 5: 13.25, 6: 25.06, 7: 0.0, 8: 12.05, 9: 0.96}, 'Apple_p': {0: 14.24, 1: 16.52, 2: 17.42, 3: 1.82, 4: 3.33, 5: 12.27, 6: 6.21, 7: 0.45, 8: 20.0, 9: 7.73}})
My code:
treeName = ['Alder','Alderm','Apple']
treeName_p = ['Alder_p','Alderm_p','Apple_p']
colornames = named.__all__
colornames = colornames[:len(treeName)]
# Create an empty figure
p = figure(x_range = temp['bydelsnavn'].values,plot_width = 700, plot_height=400,
title='Tree pr. district', toolbar_sticky = False,
tools = 'pan,wheel_zoom,reset')
# Stacked bar chart
renderers = p.vbar_stack(stackers=treeName,x='bydelsnavn',source=temp,
width=0.8, color = colornames)
# Add the hover tool
for r in renderers:
tree = r.name
hover = HoverTool(tooltips=[
("%s" % tree, "#{%s}" % tree)
], renderers = [r])
p.add_tools(hover)
# remove the grid
p.xgrid.grid_line_color=None
p.ygrid.grid_line_color=None
# Make sure bars stat at 0
p.y_range.start = 0
# remove - y-axis
p.yaxis.visible = False
# Remove the grey box around the plot
p.outline_line_color = None
# Turn the x-labels
p.xaxis.major_label_orientation = 0.5
# Remove tool bar logo
p.toolbar.logo = None
# Move the border of the left side to show "Amager"
p.min_border_left = 30
show(p)
My current chart looks like this:

Assuming that the values of the _p columns are actually in the data source, you can just add another tooltip to the HoverTool:
for r in renderers:
tree = r.name
p.add_tools(HoverTool(tooltips=[(tree, "#$name"),
(f"{tree} %", f"#{tree}_p")],
renderers=[r]))
Notice how #$name is used in there - not that necessary in this particular case but sometimes comes in handy.

Bokeh: remove Hovertool from toolbar in stacked bar chart

In my stacked barchart I have specified which tools I want in the toolbar. But when I add a hovertool this seems to overwrite my command and adds a hovertool for every element in the stacked barchart. How can I remove the hovertool tool in the toolbar?
Data example:
temp=pd.DataFrame( {'bydelsnavn': {0: 'Amager Vest', 1: 'Amager Øst', 2: 'Bispebjerg', 3: 'Brønshøj-Husum', 4: 'Indre By', 5: 'Nørrebro', 6: 'Valby', 7: 'Vanløse', 8: 'Vesterbro', 9: 'Østerbro'}, 'Alder': {0: 53.0, 1: 21.0, 2: 1.0, 3: 9.0, 4: 4.0, 5: 2.0, 6: 3.0, 7: 44.0, 8: 46.0, 9: 59.0}, 'Alderm': {0: 63.0, 1: 32.0, 2: 49.0, 3: 13.0, 4: 45.0, 5: 55.0, 6: 104.0, 7: 0.0, 8: 50.0, 9: 4.0}, 'Apple': {0: 94.0, 1: 109.0, 2: 115.0, 3: 12.0, 4: 22.0, 5: 81.0, 6: 41.0, 7: 3.0, 8: 132.0, 9: 51.0}, 'Alder_p': {0: 21.9, 1: 8.68, 2: 0.41, 3: 3.72, 4: 1.65, 5: 0.83, 6: 1.24, 7: 18.18, 8: 19.01, 9: 24.38}, 'Alderm_p': {0: 15.18, 1: 7.71, 2: 11.81, 3: 3.13, 4: 10.84, 5: 13.25, 6: 25.06, 7: 0.0, 8: 12.05, 9: 0.96}, 'Apple_p': {0: 14.24, 1: 16.52, 2: 17.42, 3: 1.82, 4: 3.33, 5: 12.27, 6: 6.21, 7: 0.45, 8: 20.0, 9: 7.73}})
My code:
treeName = ['Alder','Alderm','Apple']
treeName_p = ['Alder_p','Alderm_p','Apple_p']
colornames = named.__all__
colornames = colornames[:len(treeName)]
# Create an empty figure
p = figure(x_range = temp['bydelsnavn'].values,plot_width = 700, plot_height=400,
title='Tree pr. district', toolbar_sticky = False,
tools = 'pan,wheel_zoom,reset')
# Stacked bar chart
renderers = p.vbar_stack(stackers=treeName,x='bydelsnavn',source=temp,
width=0.8, color = colornames)
# Add the hover tool
for r in renderers:
tree = r.name
hover = HoverTool(tooltips=[
("%s" % tree, "#{%s}" % tree)
], renderers = [r])
p.add_tools(hover)
# remove the grid
p.xgrid.grid_line_color=None
p.ygrid.grid_line_color=None
# Make sure bars stat at 0
p.y_range.start = 0
# remove - y-axis
p.yaxis.visible = False
# Remove the grey box around the plot
p.outline_line_color = None
# Turn the x-labels
p.xaxis.major_label_orientation = 0.5
# Remove tool bar logo
p.toolbar.logo = None
# Move the border of the left side to show "Amager"
p.min_border_left = 30
show(p)
In the picture the tools I would like to aviod are pointed out:

You can hide the tool buttons completely by passing toggleable=False to HoverTool.

Number of labels does not match samples on decision tree regression

Trying to run a decision tree regressor on my data, but whenever I try and run my code, I get this error
ValueError: Number of labels=78177 does not match number of samples=312706
#feature selection
from sklearn.model_selection import train_test_split
from sklearn.tree import DecisionTreeRegressor
target = ['sale_price']
train, test = train_test_split(housing_data, test_size=0.2)
regression_tree = DecisionTreeRegressor(criterion="entropy",random_state=100,
max_depth=4,min_samples_leaf=5)
regression_tree.fit(train,test)
I have added a sample of my code, hopefully this gives you more context to help better understand my question and problem:
{'Age of House at Sale': {0: 6,
1: 2016,
2: 92,
3: 42,
4: 90,
5: 2012,
6: 89,
7: 3,
8: 2015,
9: 104},
'AreaSource': {0: 2.0,
1: 7.0,
2: 2.0,
3: 2.0,
4: 2.0,
5: 2.0,
6: 2.0,
7: 2.0,
8: 2.0,
9: 2.0},
'AssessLand': {0: 9900.0,
1: 1571850.0,
2: 1548000.0,
3: 36532350.0,
4: 2250000.0,
5: 3110400.0,
6: 2448000.0,
7: 1354500.0,
8: 1699200.0,
9: 1282500.0},
'AssessTot': {0: 34380.0,
1: 1571850.0,
2: 25463250.0,
3: 149792400.0,
4: 27166050.0,
5: 5579990.0,
6: 28309500.0,
7: 23965650.0,
8: 3534300.0,
9: 11295000.0},
'BldgArea': {0: 2688.0,
1: 0.0,
2: 304650.0,
3: 2548000.0,
4: 356000.0,
5: 382746.0,
6: 290440.0,
7: 241764.0,
8: 463427.0,
9: 547000.0},
'BldgClass': {0: 72,
1: 89,
2: 80,
3: 157,
4: 150,
5: 44,
6: 92,
7: 43,
8: 39,
9: 61},
'BldgDepth': {0: 50.0,
1: 0.0,
2: 92.0,
3: 0.0,
4: 100.33,
5: 315.0,
6: 125.0,
7: 100.0,
8: 0.0,
9: 80.92},
'BldgFront': {0: 20.0,
1: 0.0,
2: 335.0,
3: 0.0,
4: 202.0,
5: 179.0,
6: 92.0,
7: 500.0,
8: 0.0,
9: 304.0},
'BsmtCode': {0: 5.0,
1: 5.0,
2: 5.0,
3: 5.0,
4: 2.0,
5: 5.0,
6: 2.0,
7: 2.0,
8: 5.0,
9: 5.0},
'CD': {0: 310.0,
1: 302.0,
2: 302.0,
3: 318.0,
4: 302.0,
5: 301.0,
6: 302.0,
7: 301.0,
8: 301.0,
9: 302.0},
'ComArea': {0: 0.0,
1: 0.0,
2: 304650.0,
3: 2548000.0,
4: 30000.0,
5: 11200.0,
6: 290440.0,
7: 27900.0,
8: 4884.0,
9: 547000.0},
'CommFAR': {0: 0.0,
1: 2.0,
2: 2.0,
3: 2.0,
4: 0.0,
5: 0.0,
6: 10.0,
7: 2.0,
8: 0.0,
9: 2.0},
'Council': {0: 41.0,
1: 33.0,
2: 33.0,
3: 46.0,
4: 33.0,
5: 33.0,
6: 33.0,
7: 33.0,
8: 33.0,
9: 35.0},
'Easements': {0: 0.0,
1: 0.0,
2: 0.0,
3: 1.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 0.0,
8: 0.0,
9: 0.0},
'ExemptLand': {0: 0.0,
1: 1571850.0,
2: 0.0,
3: 0.0,
4: 2250000.0,
5: 0.0,
6: 0.0,
7: 932847.0,
8: 0.0,
9: 0.0},
'ExemptTot': {0: 0.0,
1: 1571850.0,
2: 0.0,
3: 0.0,
4: 27166050.0,
5: 0.0,
6: 11304900.0,
7: 23543997.0,
8: 0.0,
9: 0.0},
'FacilFAR': {0: 0.0,
1: 6.5,
2: 0.0,
3: 0.0,
4: 4.8,
5: 4.8,
6: 10.0,
7: 3.0,
8: 5.0,
9: 4.8},
'FactryArea': {0: 0.0,
1: 0.0,
2: 0.0,
3: 0.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 0.0,
8: 0.0,
9: 547000.0},
'GarageArea': {0: 0.0,
1: 0.0,
2: 0.0,
3: 1285000.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 22200.0,
8: 0.0,
9: 0.0},
'HealthArea': {0: 6410.0,
1: 1000.0,
2: 2300.0,
3: 8822.0,
4: 2300.0,
5: 400.0,
6: 2300.0,
7: 700.0,
8: 500.0,
9: 9300.0},
'HealthCent': {0: 35.0,
1: 36.0,
2: 38.0,
3: 35.0,
4: 38.0,
5: 30.0,
6: 38.0,
7: 30.0,
8: 30.0,
9: 36.0},
'IrrLotCode': {0: 1, 1: 1, 2: 0, 3: 0, 4: 1, 5: 1, 6: 0, 7: 1, 8: 0, 9: 0},
'LandUse': {0: 2.0,
1: 10.0,
2: 5.0,
3: 5.0,
4: 8.0,
5: 4.0,
6: 5.0,
7: 3.0,
8: 3.0,
9: 6.0},
'LotArea': {0: 2252.0,
1: 134988.0,
2: 32000.0,
3: 905000.0,
4: 20267.0,
5: 57600.0,
6: 12500.0,
7: 50173.0,
8: 44704.0,
9: 113800.0},
'LotDepth': {0: 100.0,
1: 275.33,
2: 335.92,
3: 859.0,
4: 100.33,
5: 320.0,
6: 125.0,
7: 200.0,
8: 281.86,
9: 204.0},
'LotFront': {0: 24.0,
1: 490.5,
2: 92.42,
3: 930.0,
4: 202.0,
5: 180.0,
6: 100.0,
7: 521.25,
8: 225.08,
9: 569.0},
'LotType': {0: 5.0,
1: 5.0,
2: 3.0,
3: 3.0,
4: 3.0,
5: 3.0,
6: 3.0,
7: 1.0,
8: 5.0,
9: 3.0},
'NumBldgs': {0: 1.0,
1: 0.0,
2: 1.0,
3: 4.0,
4: 1.0,
5: 1.0,
6: 1.0,
7: 1.0,
8: 2.0,
9: 13.0},
'NumFloors': {0: 2.0,
1: 0.0,
2: 13.0,
3: 2.0,
4: 15.0,
5: 0.0,
6: 37.0,
7: 6.0,
8: 20.0,
9: 8.0},
'OfficeArea': {0: 0.0,
1: 0.0,
2: 264750.0,
3: 0.0,
4: 30000.0,
5: 1822.0,
6: 274500.0,
7: 4200.0,
8: 0.0,
9: 0.0},
'OtherArea': {0: 0.0,
1: 0.0,
2: 39900.0,
3: 0.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 0.0,
8: 0.0,
9: 0.0},
'PolicePrct': {0: 70.0,
1: 84.0,
2: 84.0,
3: 63.0,
4: 84.0,
5: 90.0,
6: 84.0,
7: 94.0,
8: 90.0,
9: 88.0},
'ProxCode': {0: 0.0,
1: 0.0,
2: 0.0,
3: 0.0,
4: 0.0,
5: 0.0,
6: 0.0,
7: 1.0,
8: 0.0,
9: 0.0},
'ResArea': {0: 2172.0,
1: 0.0,
2: 0.0,
3: 0.0,
4: 0.0,
5: 371546.0,
6: 0.0,
7: 213864.0,
8: 458543.0,
9: 0.0},
'ResidFAR': {0: 2.0,
1: 7.2,
2: 0.0,
3: 0.0,
4: 2.43,
5: 2.43,
6: 10.0,
7: 3.0,
8: 5.0,
9: 0.0},
'RetailArea': {0: 0.0,
1: 0.0,
2: 0.0,
3: 1263000.0,
4: 0.0,
5: 9378.0,
6: 15940.0,
7: 0.0,
8: 4884.0,
9: 0.0},
'SHAPE_Area': {0: 2316.8863224,
1: 140131.577176,
2: 34656.4472405,
3: 797554.847834,
4: 21360.1476315,
5: 58564.8643115,
6: 12947.145471,
7: 50772.624868800005,
8: 47019.5677861,
9: 118754.78573699998},
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'SplitZone': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 0, 9: 1},
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'UnitsRes': {0: 2.0,
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'UnitsTotal': {0: 2.0,
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'YearAlter1': {0: 0.0,
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'ZoneMap': {0: 3,
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'building_class': {0: 141,
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'building_class_at_sale': {0: 143,
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'gross_sqft': {0: 0.0,
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'tax_class': {0: 3, 1: 3, 2: 3, 3: 3, 4: 3, 5: 3, 6: 3, 7: 7, 8: 3, 9: 3},
'total_units': {0: 1,
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9: 11205}}

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