import pandas as pd
import numpy as np
%pylab inline

Populating the interactive namespace from numpy and matplotlib

data = pd.read_csv("data_1304.csv",encoding='latin-1')

data.dropna(inplace=True)

pd.to_datetime(data['_started_at'].head())

0   2018-01-10 15:55:00
1   2018-01-10 17:04:00
2   2018-01-10 15:53:00
3   2018-01-11 02:24:00
4   2018-01-11 05:14:00
Name: _started_at, dtype: datetime64[ns]

Time spent on a question (can be useful for worker ability)

pd.to_datetime(data['_created_at'])- pd.to_datetime(data['_started_at']) #use pd.to_numeric() to convert to number of ns

0    00:25:00
1    00:19:20
2    00:28:06
4    00:07:22
5    00:25:10
6    00:13:59
7    00:16:48
8    00:16:12
9    00:22:55
10   00:13:19
11   00:10:26
13   00:17:22
14   00:17:55
15   00:24:56
16   00:27:48
17   00:25:27
18   00:28:41
19   00:22:18
20   00:08:38
21   00:06:07
22   00:28:04
23   00:27:28
25   00:23:54
26   00:20:47
27   00:18:40
29   00:09:51
30   00:17:40
33   00:09:06
34   00:24:43
35   00:27:52
       ...   
57   00:26:12
59   00:13:00
61   00:21:06
62   00:24:37
63   00:25:15
65   00:15:52
66   00:12:22
68   00:14:17
70   00:24:54
71   00:29:49
72   00:23:56
74   00:24:29
75   00:28:43
76   00:16:54
77   00:20:06
78   00:24:14
79   00:20:13
82   00:24:47
83   00:18:22
85   00:18:31
87   00:21:30
88   00:20:33
89   00:23:10
93   00:26:09
94   00:20:54
95   00:19:57
96   00:29:33
97   00:23:53
98   00:21:24
99   00:27:11
Length: 73, dtype: timedelta64[ns]

print(len(data))
data

100

	better_0	_unit_id	_started_at	_created_at	_trust	_worker_id	_city	age	similarity_0	explanation_0	asi1
0	Your keyword	4	01.10.18 15:55	01.10.18 16:20	0.385118	32	Ernakulam	36-50	6	they are all dressed well and using computers ...	4
1	The two keywords are completely identical	6	01.10.18 17:04	01.10.18 17:23	0.033270	13	Kolkata	36-50	6	Almost identicalexcept the tiny spelling diffe...	3
2	Search engine query	18	01.10.18 15:53	01.10.18 16:21	0.551213	21	Pune	36-50	6	All the images represents the search better...	1
3	The two keywords are completely identical	1	01.11.18 02:24	01.11.18 02:47	0.204184	82	NaN	19-25	7	they wear casual clothes	4
4	The two keywords are completely identical	6	01.11.18 05:14	01.11.18 05:21	0.708808	70	Mangalagiri	19-25	7	both are similar	5
5	The two keywords are completely identical	20	01.10.18 16:41	01.10.18 17:06	0.899786	95	Patna	26-35	7	they both describe the same kind of people	3
6	Your keyword	13	01.10.18 15:47	01.10.18 16:01	0.873825	37	Ulhasnagar	19-25	6	We can see a relaxed state in that images	5
7	Search engine query	30	01.10.18 15:40	01.10.18 15:57	0.264847	78	Kolkata	19-25	5	YES	3
8	Your keyword	9	01.11.18 04:55	01.11.18 05:11	0.512431	29	Mangalagiri	19-25	6	THEY ARE THINKING	3
9	Your keyword	7	01.10.18 15:39	01.10.18 16:02	0.260237	87	Hyderabad	26-35	6	A person is generalized and one cannot find th...	5
10	Search engine query	10	01.10.18 16:25	01.10.18 16:38	0.915093	91	New Delhi	19-25	4	they are calm	3
11	Your keyword	25	01.11.18 03:56	01.11.18 04:06	0.212509	79	Roorkee	19-25	4	genious	3
12	The two keywords are completely identical	17	01.10.18 17:01	01.10.18 17:26	0.557112	43	Chennai	26-35	6	NaN	3
13	Search engine query	10	01.10.18 19:57	01.10.18 20:15	0.770169	6	Hyderabad	26-35	6	only 1 image	4
14	The two keywords are completely identical	10	01.11.18 03:02	01.11.18 03:20	0.914456	86	Roorkee	26-35	7	interested in their work	4
15	Search engine query	26	01.10.18 16:49	01.10.18 17:14	0.283502	71	Cochin	19-25	3	i think this is correct that calm person becau...	3
16	Search engine query	4	01.10.18 20:14	01.10.18 20:42	0.373995	64	Kolkata	19-25	5	YES	4
17	Your keyword	14	01.10.18 17:45	01.10.18 18:11	0.035891	85	Mumbai	26-35	4	images looks like taking a deep breath	5
18	Search engine query	14	01.11.18 04:34	01.11.18 05:02	0.797305	34	Amritsar	26-35	4	it now seems more like to give these results w...	4
19	Your keyword	15	01.10.18 19:23	01.10.18 19:45	0.814008	11	Bhopal	36-50	4	based on result of image	5
20	Your keyword	30	01.11.18 04:06	01.11.18 04:15	0.634484	80	Dehradun	19-25	4	whipping	3
21	Search engine query	17	01.10.18 16:51	01.10.18 16:57	0.613763	98	New Delhi	19-25	4	Yes	2
22	The two keywords are completely identical	10	01.10.18 16:19	01.10.18 16:47	0.189142	17	Kolkata	36-50	7	calm person and calmness same	5
23	Your keyword	1	01.10.18 16:49	01.10.18 17:17	0.801677	20	Bangalore	26-35	5	result suits more to this kerword	4
24	Your keyword	27	01.11.18 04:23	01.11.18 04:47	0.825728	84	NaN	19-25	4	engry	2
25	Search engine query	28	01.10.18 15:44	01.10.18 16:08	0.935156	83	Pune	19-25	5	working person uses the things that i mentioned	4
26	Search engine query	15	01.10.18 16:23	01.10.18 16:44	0.851697	94	Kolkata	19-25	6	yes	3
27	The two keywords are completely identical	27	01.10.18 17:12	01.10.18 17:30	0.796527	65	Meerut	19-25	4	anger	2
28	The two keywords are completely identical	9	01.11.18 05:12	01.11.18 05:25	0.333228	61	Bokaro	19-25	6	NaN	4
29	Search engine query	16	01.10.18 16:11	01.10.18 16:21	0.940862	90	Dehradun	36-50	4	hot air baloon	3
...	...	...	...	...	...	...	...	...	...	...	...
70	Your keyword	17	01.11.18 04:26	01.11.18 04:51	0.427894	53	Mangalagiri	19-25	6	by query image i understood that person seems ...	3
71	Your keyword	7	01.10.18 23:35	01.11.18 00:04	0.621600	28	Kolkata	50-80	7	Everything is related with warm	3
72	Search engine query	7	01.10.18 16:06	01.10.18 16:30	0.677330	38	Hyderabad	26-35	5	it gives better ideas about all the image	4
73	Search engine query	11	01.11.18 02:38	01.11.18 02:54	0.504001	56	Chennai	19-25	5	NaN	4
74	Search engine query	24	01.11.18 04:06	01.11.18 04:31	0.354657	67	Mangalagiri	19-25	7	we got the same image when search in google	5
75	Search engine query	16	01.10.18 19:32	01.10.18 20:01	0.090422	52	Hyderabad	26-35	6	with the facial expression we can find him too...	4
76	Search engine query	19	01.10.18 16:31	01.10.18 16:47	0.967048	59	Pune	19-25	6	Their some people shouting at each other	4
77	Your keyword	2	01.10.18 21:04	01.10.18 21:25	0.928919	25	Howrah	19-25	6	very much about that	3
78	Search engine query	24	01.10.18 16:36	01.10.18 17:00	0.525274	77	Kolkata	26-35	7	It's the image of that	3
79	The two keywords are completely identical	21	01.11.18 03:08	01.11.18 03:28	0.588048	31	Unnao	26-35	7	Smart Person Bring Innovation and must have hi...	0
80	The two keywords are completely identical	31	01.10.18 15:43	01.10.18 16:03	0.916103	41	Nellore	26-35	7	NaN	4
81	Search engine query	30	01.10.18 17:14	01.10.18 17:34	0.752203	88	NaN	NaN	6	it is more relative.	5
82	The two keywords are completely identical	4	01.10.18 15:39	01.10.18 16:04	0.625525	3	Hyderabad	36-50	5	person in aggression is shouting at others	5
83	Your keyword	30	01.11.18 05:56	01.11.18 06:15	0.042048	5	Burdwan	26-35	7	they are all were casual dress	5
84	The two keywords are completely identical	23	01.11.18 03:52	01.11.18 04:07	0.464201	39	Siuri	26-35	7	NaN	5
85	Your keyword	7	01.10.18 17:08	01.10.18 17:26	0.280744	49	Noida	0-18	3	By nature	5
86	Search engine query	29	01.10.18 16:13	01.10.18 16:39	0.278253	73	NaN	26-35	7	Casual suits more than free style	4
87	Your keyword	24	01.10.18 19:54	01.10.18 20:15	0.888706	35	Delhi	26-35	7	because it shows that	3
88	The two keywords are completely identical	27	01.11.18 04:05	01.11.18 04:26	0.058092	55	Mangalagiri	19-25	7	we got the same image when search in google	3
89	Search engine query	21	01.10.18 15:50	01.10.18 16:13	0.805073	50	Hyderabad	26-35	7	people are working i guess working people is m...	4
90	Your keyword	20	01.10.18 15:50	01.10.18 16:01	0.761000	96	NaN	26-35	4	frustreted	2
91	Your keyword	17	01.11.18 03:56	01.11.18 04:21	0.117168	9	Erode	26-35	6	NaN	4
92	The two keywords are completely identical	21	01.10.18 16:11	01.10.18 16:22	0.343565	93	Pune	26-35	7	NaN	5
93	The two keywords are completely identical	3	01.11.18 04:13	01.11.18 04:39	0.853188	45	Mangalagiri	19-25	7	BOTH ARE SIMILAR	5
94	Your keyword	13	01.10.18 16:01	01.10.18 16:22	0.325484	57	Guwahati	36-50	6	They also look happy	4
95	The two keywords are completely identical	11	01.11.18 05:45	01.11.18 06:05	0.988551	23	Mangalagiri	0-18	7	similar	5
96	Search engine query	8	01.11.18 02:23	01.11.18 02:52	0.520720	0	Hyderabad	26-35	5	Since not all the images belong to science exa...	4
97	Search engine query	15	01.11.18 01:56	01.11.18 02:20	0.046097	7	Chennai	26-35	5	On detailed viewing smart person might be a be...	4
98	Your keyword	3	01.11.18 06:49	01.11.18 07:11	0.091185	27	Kolkata	36-50	5	everybody is yelling	4
99	Search engine query	31	01.11.18 04:19	01.11.18 04:46	0.951531	16	Mangalagiri	19-25	4	a complete act of expression works out here	3

100 rows × 11 columns

data.describe()

	_unit_id	_trust	_worker_id	similarity_0	asi1
count	100.000000	100.000000	100.000000	100.000000	100.000000
mean	15.730000	0.542089	49.500000	5.640000	3.680000
std	8.539693	0.301171	29.011492	1.275408	1.071957
min	0.000000	0.033270	0.000000	2.000000	0.000000
25%	9.000000	0.295970	24.750000	5.000000	3.000000
50%	15.000000	0.571722	49.500000	6.000000	4.000000
75%	23.000000	0.807306	74.250000	7.000000	4.000000
max	31.000000	0.988551	99.000000	7.000000	5.000000

Let's see how many judgments we have per unit

data.groupby('better_0').size()

better_0
Search engine query                          33
The two keywords are completely identical    32
Your keyword                                 35
dtype: int64

data.groupby('_unit_id').size()

array([1, 3, 3, 2, 4, 1, 2, 7, 1, 2, 4, 3, 2, 4, 6, 7, 3, 5, 3, 1, 2, 4,
       1, 6, 5, 3, 2, 4, 2, 1, 4, 2])

data.groupby('_unit_id').size().hist()

<matplotlib.axes._subplots.AxesSubplot at 0x7ff0abee0d30>

Let's remove the units that have only one judgment

(data.groupby('_unit_id').size()==1).values

array([ True, False, False, False, False,  True, False, False,  True,
        True, False,  True, False, False, False, False, False, False,
        True,  True, False, False, False, False, False, False, False,
       False,  True])

a = np.where((data.groupby('_unit_id').size()==1))
a

(array([ 0,  5,  8,  9, 11, 18, 19, 28]),)

a = list(a[0])
a

[0, 5, 8, 9, 11, 18, 19, 28]

data[data['_unit_id'].isin(a)]

data = data[~data['_unit_id'].isin(a)]
len(data)

63

1. Create a column with time spent (use pd.to_datetime)
2. Compute the average time per worker

data['time_spent'] = pd.to_datetime(data['_created_at']) - pd.to_datetime(data['_started_at'])
data.head()

	better_0	_unit_id	_started_at	_created_at	_trust	_worker_id	_city	age	similarity_0	explanation_0	asi1	time_spent
0	Your keyword	4	1/10/2018 15:55:41	1/10/2018 16:20:41	0.385118	32	Ernakulam	36-50	6	they are all dressed well and using computers ...	4	00:25:00
1	The two keywords are completely identical	6	1/10/2018 17:04:22	1/10/2018 17:23:42	0.033270	13	Kolkata	36-50	6	Almost identicalexcept the tiny spelling diffe...	3	00:19:20
4	The two keywords are completely identical	6	1/11/2018 05:14:03	1/11/2018 05:21:25	0.708808	70	Mangalagiri	19-25	7	both are similar	5	00:07:22
5	The two keywords are completely identical	20	1/10/2018 16:41:16	1/10/2018 17:06:26	0.899786	95	Patna	26-35	7	they both describe the same kind of people	3	00:25:10
6	Your keyword	13	1/10/2018 15:47:20	1/10/2018 16:01:19	0.873825	37	Ulhasnagar	19-25	6	We can see a relaxed state in that images	5	00:13:59

Basic aggregation

Quantitative variables

data.groupby('_unit_id')['similarity_0'].mean()

_unit_id
1     5.500000
2     4.666667
3     6.000000
4     5.750000
6     6.500000
7     5.400000
10    6.000000
13    6.333333
14    4.000000
15    5.166667
16    5.000000
17    5.666667
20    7.000000
21    7.000000
23    6.500000
24    6.200000
25    4.333333
26    4.500000
27    5.666667
30    5.333333
31    4.000000
Name: similarity_0, dtype: float64

If we are also doing a per-worker analysis, we can compute values from the worker

data.groupby('_worker_id')['_trust'].mean().values

array([0.45086904, 0.92770687, 0.62552536, 0.93464997, 0.04204837,
       0.77016858, 0.04609719, 0.60929146, 0.5741613 , 0.81400838,
       0.03326987, 0.91541507, 0.95153081, 0.18914215, 0.30560117,
       0.80167709, 0.97441615, 0.92891881, 0.96747946, 0.09118499,
       0.62159951, 0.58563959, 0.58804797, 0.38511825, 0.12424342,
       0.79730475, 0.88870635, 0.87382468, 0.67732971, 0.85318828,
       0.34576951, 0.28074398, 0.80507253, 0.05786407, 0.09042158,
       0.42789365, 0.05809224, 0.32548398, 0.30012607, 0.03610733,
       0.85113121, 0.37399525, 0.79652694, 0.62465149, 0.3546574 ,
       0.91410825, 0.70880836, 0.28350176, 0.91083596, 0.33243423,
       0.03891988, 0.52527424, 0.26484709, 0.21250903, 0.63448413,
       0.03589091, 0.91445626, 0.2602371 , 0.9408621 , 0.9150934 ,
       0.85169676, 0.89978581, 0.61376345])

data.groupby('_worker_id')['_trust'].mean().hist()

<matplotlib.axes._subplots.AxesSubplot at 0x7ff0b3eff390>

Categorical variables

Now we can't do the following because the following is a categorical variable:

data.groupby('_unit_id')['better_0'].mean()

---------------------------------------------------------------------------
DataError                                 Traceback (most recent call last)
<ipython-input-20-2f584eae24bb> in <module>()
----> 1 data.groupby('_unit_id')['better_0'].mean()

/srv/paws/lib/python3.6/site-packages/pandas/core/groupby.py in mean(self, *args, **kwargs)
   1126         nv.validate_groupby_func('mean', args, kwargs, ['numeric_only'])
   1127         try:
-> 1128             return self._cython_agg_general('mean', **kwargs)
   1129         except GroupByError:
   1130             raise

/srv/paws/lib/python3.6/site-packages/pandas/core/groupby.py in _cython_agg_general(self, how, alt, numeric_only, min_count)
    925 
    926         if len(output) == 0:
--> 927             raise DataError('No numeric types to aggregate')
    928 
    929         return self._wrap_aggregated_output(output, names)

DataError: No numeric types to aggregate

Let's explore what is this column and decide what to do

data.groupby('_unit_id')['better_0'].describe() 

	count	unique	top	freq
_unit_id
1	2	2	Your keyword	1
2	3	2	Your keyword	2
3	2	2	Your keyword	1
4	4	3	The two keywords are completely identical	2
6	2	1	The two keywords are completely identical	2
7	5	2	Your keyword	3
10	4	2	The two keywords are completely identical	2
13	3	2	Your keyword	2
14	3	2	Search engine query	2
15	6	2	Your keyword	4
16	2	1	Search engine query	2
17	3	3	Your keyword	1
20	1	1	The two keywords are completely identical	1
21	2	2	Search engine query	1
23	4	1	The two keywords are completely identical	4
24	5	2	Search engine query	3
25	3	3	Your keyword	1
26	2	1	Search engine query	2
27	3	2	The two keywords are completely identical	2
30	3	2	Your keyword	2
31	1	1	Search engine query	1

print(data['better_0'].unique())
len(data['better_0'].unique())

['Your keyword' 'The two keywords are completely identical'
 'Search engine query']

3

The majority vote of an array is simply the mode

data['better_0'].mode()

0    Search engine query
1           Your keyword
dtype: object

How is the variable distributed?

data.groupby('better_0')['better_0'].size()

better_0
Search engine query                          22
The two keywords are completely identical    19
Your keyword                                 22
Name: better_0, dtype: int64

Let's compute the majority voting

data.groupby('_unit_id')['better_0'].apply(lambda x: x.mode())

_unit_id   
1         0    The two keywords are completely identical
          1                                 Your keyword
2         0                                 Your keyword
3         0    The two keywords are completely identical
          1                                 Your keyword
4         0    The two keywords are completely identical
6         0    The two keywords are completely identical
7         0                                 Your keyword
10        0                          Search engine query
          1    The two keywords are completely identical
13        0                                 Your keyword
14        0                          Search engine query
15        0                                 Your keyword
16        0                          Search engine query
17        0                          Search engine query
          1    The two keywords are completely identical
          2                                 Your keyword
20        0    The two keywords are completely identical
21        0                          Search engine query
          1    The two keywords are completely identical
23        0    The two keywords are completely identical
24        0                          Search engine query
25        0                          Search engine query
          1    The two keywords are completely identical
          2                                 Your keyword
26        0                          Search engine query
27        0    The two keywords are completely identical
30        0                                 Your keyword
31        0                          Search engine query
Name: better_0, dtype: object

Sometimes this returns two values, let's get the first in that case (better way would be random)

data.groupby('_unit_id')['better_0'].apply(lambda x: x.mode()[0])

_unit_id
1     The two keywords are completely identical
2                                  Your keyword
3     The two keywords are completely identical
4     The two keywords are completely identical
6     The two keywords are completely identical
7                                  Your keyword
10                          Search engine query
13                                 Your keyword
14                          Search engine query
15                                 Your keyword
16                          Search engine query
17                          Search engine query
20    The two keywords are completely identical
21                          Search engine query
23    The two keywords are completely identical
24                          Search engine query
25                          Search engine query
26                          Search engine query
27    The two keywords are completely identical
30                                 Your keyword
31                          Search engine query
Name: better_0, dtype: object

Weighted measures

Weighted mean

def weigthed_mean(df,weights,values): #df is a dataframe containing a single question
    sum_values = (df[weights]*df[values]).sum()
    total_weight = df[weights].sum()
    return sum_values/total_weight

data.groupby('_unit_id').apply(lambda x: weigthed_mean(x,'_trust','similarity_0'))

_unit_id
1     5.532764
2     4.961362
3     6.806888
4     5.789938
6     6.955167
7     5.675547
10    5.739468
13    6.437989
14    4.000000
15    5.166934
16    4.175357
17    5.840521
20    7.000000
21    7.000000
23    6.465985
24    6.481138
25    4.525120
26    4.556271
27    4.706914
30    4.415340
31    4.000000
dtype: float64

data.groupby('_unit_id').apply(lambda x: (x['_trust']*x['similarity_0']).sum()/(x['_trust'].sum()))

_unit_id
1     5.532764
2     4.961362
3     6.806888
4     5.789938
6     6.955167
7     5.675547
10    5.739468
13    6.437989
14    4.000000
15    5.166934
16    4.175357
17    5.840521
20    7.000000
21    7.000000
23    6.465985
24    6.481138
25    4.525120
26    4.556271
27    4.706914
30    4.415340
31    4.000000
dtype: float64

Weighted majority voting

Now we need, for each unit, to find the category with the highest trust score

data.head()

	better_0	_unit_id	_started_at	_created_at	_trust	_worker_id	_city	age	similarity_0	explanation_0	asi1	time_spent
0	Your keyword	4	1/10/2018 15:55:41	1/10/2018 16:20:41	0.385118	32	Ernakulam	36-50	6	they are all dressed well and using computers ...	4	00:25:00
1	The two keywords are completely identical	6	1/10/2018 17:04:22	1/10/2018 17:23:42	0.033270	13	Kolkata	36-50	6	Almost identicalexcept the tiny spelling diffe...	3	00:19:20
4	The two keywords are completely identical	6	1/11/2018 05:14:03	1/11/2018 05:21:25	0.708808	70	Mangalagiri	19-25	7	both are similar	5	00:07:22
5	The two keywords are completely identical	20	1/10/2018 16:41:16	1/10/2018 17:06:26	0.899786	95	Patna	26-35	7	they both describe the same kind of people	3	00:25:10
6	Your keyword	13	1/10/2018 15:47:20	1/10/2018 16:01:19	0.873825	37	Ulhasnagar	19-25	6	We can see a relaxed state in that images	5	00:13:59

def weigthed_majority(df,weights,values): #df is a dataframe containing a single question
    #print(df.groupby(values)[weights].sum())
    best_value = df.groupby(values)[weights].sum().argmax()
    return best_value

data.groupby('_unit_id').apply(lambda x: weigthed_majority(x,'_trust','better_0'))

/srv/paws/lib/python3.6/site-packages/ipykernel_launcher.py:3: FutureWarning: 'argmax' is deprecated. Use 'idxmax' instead. The behavior of 'argmax' will be corrected to return the positional maximum in the future. Use 'series.values.argmax' to get the position of the maximum now.
  This is separate from the ipykernel package so we can avoid doing imports until

_unit_id
1     The two keywords are completely identical
2                                  Your keyword
3     The two keywords are completely identical
4     The two keywords are completely identical
6     The two keywords are completely identical
7                           Search engine query
10                          Search engine query
13                                 Your keyword
14                          Search engine query
15                                 Your keyword
16                          Search engine query
17    The two keywords are completely identical
20    The two keywords are completely identical
21                          Search engine query
23    The two keywords are completely identical
24                          Search engine query
25    The two keywords are completely identical
26                          Search engine query
27    The two keywords are completely identical
30                                 Your keyword
31                          Search engine query
dtype: object

Creating a summary table

results = pd.DataFrame()
results['better'] = data.groupby('_unit_id').apply(lambda x: weigthed_majority(x,'_trust','better_0'))
results['similarity'] = data.groupby('_unit_id').apply(lambda x: weigthed_mean(x,'_trust','similarity_0'))
results['better_code'] = results['better'].astype('category').cat.codes
results

/srv/paws/lib/python3.6/site-packages/ipykernel_launcher.py:3: FutureWarning: 'argmax' is deprecated. Use 'idxmax' instead. The behavior of 'argmax' will be corrected to return the positional maximum in the future. Use 'series.values.argmax' to get the position of the maximum now.
  This is separate from the ipykernel package so we can avoid doing imports until

	better	similarity	better_code
_unit_id
1	The two keywords are completely identical	5.532764	1
2	Your keyword	4.961362	2
3	The two keywords are completely identical	6.806888	1
4	The two keywords are completely identical	5.789938	1
6	The two keywords are completely identical	6.955167	1
7	Search engine query	5.675547	0
10	Search engine query	5.739468	0
13	Your keyword	6.437989	2
14	Search engine query	4.000000	0
15	Your keyword	5.166934	2
16	Search engine query	4.175357	0
17	The two keywords are completely identical	5.840521	1
20	The two keywords are completely identical	7.000000	1
21	Search engine query	7.000000	0
23	The two keywords are completely identical	6.465985	1
24	Search engine query	6.481138	0
25	The two keywords are completely identical	4.525120	1
26	Search engine query	4.556271	0
27	The two keywords are completely identical	4.706914	1
30	Your keyword	4.415340	2
31	Search engine query	4.000000	0

Free text

Now we analyse the case in which we have free text

data['better_0'].unique()

array(['Your keyword', 'The two keywords are completely identical',
       'Search engine query'], dtype=object)

data['explanation_0'].unique()

array(['they are all dressed well and using computers so its more like a  business scenario.',
       'Almost identicalexcept the tiny spelling difference.',
       'both are similar', 'they both describe the same kind of people',
       'We can see a relaxed state in that images', 'YES',
       'A person is generalized and one cannot find the images of Einstein or kids in them.',
       'they are calm', 'genious', 'only 1 image',
       'interested in their work',
       'i think this is correct that calm person because every one is calm in this images',
       'images looks like taking a deep breath',
       'it now seems more like to give these results whn we think of interested person rather than thinking and surprising',
       'based on result of image', 'whipping', 'Yes',
       'calm person and calmness same',
       'result suits more to this kerword', 'yes', 'anger',
       'hot air baloon', 'both are the same', 'same attitude of boss',
       'the results are same',
       'all my words are feature of Search engine query',
       'They all are working in the office',
       'in image person looking very casual',
       'both refer to the same traits but intelligent word is more suited',
       'i know', 'Because all people here look casual.', 'both are same',
       'Casualness is used in both the words',
       'interested person only can do Research, smart, thinging',
       'Casual person is more accurate of the images.',
       'i believe this is my personal theory..so i think aggressive person would be better keyword for these images',
       'My keyword "happy people" and Search engine query "calm person" is almost same.',
       'My answer is more specific regarding images.',
       'i know need search engine when i already knew it',
       'BOTH ARE SIMILAR',
       'by query image i understood that person seems very angry',
       'Everything is related with warm',
       'it gives better ideas about all the image',
       'we got the same image when search in google',
       'with the facial expression we can find him too aggresive',
       'very much about that', "It's the image of that",
       'Smart Person Bring Innovation and must have high IQ',
       'person in aggression is shouting at others',
       'they are all were casual dress', 'By nature',
       'because it shows that',
       'people are working i guess working people is more apt',
       'They also look happy',
       'On detailed viewing smart person might be a better keyword.',
       'everybody is yelling',
       'a complete act of expression works out here'], dtype=object)

We can't use the weighted majority voting here! We need first to assign a score to this values.

Exercise

• Create a function that assigns a score to each value of the column 'explanation_0' (for example the text lenght len(text)) look here for reference https://pandas.pydata.org/pandas-docs/stable/text.html
• create a column with this score
• generate a weighted mean for it (using '_trust')

def compute_score(text):
    score = len(text)
    return score

data['score'] = data['explanation_0'].apply(compute_score)

data.groupby('_unit_id').apply(lambda x: weigthed_mean(x,'_trust','score'))

48.263201486681275

Exercise

1. aggregate per _unit_id using average means for time_spent (you need to apply pd.to_numeric() and divide by 1e9 to get a column in seconds
2. create a code that assigns 1 if the text contains any element for a list of words (list_words=['similar','same'], by doing a for loop (for i in list_words) and checking with (if i in text)
3. compute the weigthed mean for that