I have this Multi Numinal regression model done by statsmodel:
writer = pd.ExcelWriter(path=os.path.join(export_path, f'regression.xlsx'), engine='xlsxwriter')
vars_matrix_df = pd.read_csv(data_path, skipinitialspace=True)
corr_cols = ['sales_vs_service', 'agent_experience', 'minutes_passed_since_shift_started', 'stage_in_conv',
'current_cust_wait_time', 'prev_cust_line_words', 'total_cust_words_in_conv',
'agent_total_turns', 'sentiment_score', 'max_sentiment', 'min_sentiment', 'last_sentiment',
'agent_response_time', 'customer_response_rate', 'is_last_cust_answered',
'conversation_opening', 'queue_length', 'total_lines_from_rep',
'agent_number_of_conversations', 'concurrency', 'rep_shift_start_time', 'first_cust_line_num_of_words',
'queue_wait_time', 'day_of_week', 'time_of_day']
reg_equation = st.formula.mnlogit(f'visitor_was_answered ~C(day_of_week)+C(time_of_day)+{"+".join(corr_cols)} ',
vars_matrix_df).fit()
the reg results:
visitor_was_answered=1 coef std err z P>|z| \
0 C(time_of_day)[T.10] 0.0071 1910000.000 3.700000e-09 1.000
1 C(time_of_day)[T.11] 0.0067 698000.000 9.600000e-09 1.000
2 C(time_of_day)[T.12] 0.0016 1790000.000 9.200000e-10 1.000
3 C(time_of_day)[T.13] 0.0031 561000.000 5.570000e-09 1.000
4 C(time_of_day)[T.14] 0.0037 1310000.000 2.840000e-09 1.000
5 C(time_of_day)[T.15] 0.0011 548000.000 2.020000e-09 1.000
6 C(time_of_day)[T.17] 0.0044 814000.000 5.440000e-09 1.000
7 C(time_of_day)[T.18] 0.0009 1100000.000 8.270000e-10 1.000
8 C(time_of_day)[T.19] 0.0047 835000.000 5.640000e-09 1.000
9 C(time_of_day)[T.20] 0.0009 1140000.000 8.100000e-10 1.000
10 time_of_day[T.10] 0.0071 1930000.000 3.670000e-09 1.000
11 time_of_day[T.11] 0.0067 686000.000 9.770000e-09 1.000
12 time_of_day[T.12] 0.0016 1800000.000 9.150000e-10 1.000
13 time_of_day[T.13] 0.0031 556000.000 5.620000e-09 1.000
14 time_of_day[T.14] 0.0037 1240000.000 3.010000e-09 1.000
15 time_of_day[T.15] 0.0011 638000.000 1.740000e-09 1.000
16 time_of_day[T.17] 0.0044 1010000.000 4.400000e-09 1.000
17 time_of_day[T.18] 0.0009 1130000.000 8.020000e-10 1.000
18 time_of_day[T.19] 0.0047 860000.000 5.480000e-09 1.000
19 time_of_day[T.20] 0.0009 1120000.000 8.270000e-10 1.000
20 sales_vs_service -0.0448 0.006 -8.102000e+00 0.000
21 agent_experience -0.0414 0.008 -4.955000e+00 0.000
22 current_cust_wait_time -39.1333 0.414 -9.457400e+01 0.000
23 prev_cust_line_words 20.0439 0.236 8.494600e+01 0.000
24 agent_total_turns 0.1110 0.038 2.949000e+00 0.003
25 sentiment_score -4.3454 0.157 -2.759000e+01 0.000
26 agent_response_time -118.0821 2.205 -5.354600e+01 0.000
27 customer_response_rate -7.0865 0.630 -1.125500e+01 0.000
28 is_last_cust_answered -0.2537 0.005 -4.860800e+01 0.000
29 conversation_opening -0.4533 0.006 -7.206300e+01 0.000
30 queue_length -1.5427 0.018 -8.642700e+01 0.000
31 agent_number_of_conversations 0.0013 0.018 7.300000e-02 0.941
32 first_cust_line_num_of_words -3.7545 0.123 -3.056900e+01 0.000
33 queue_wait_time -0.3308 0.166 -1.997000e+00 0.046
To this regression, I want to add the odds ratio values of each variable. I think that the coefficients are already odds ratio but I didn't find any proof to that. Any idea how this can be done? and what are the coefficients represent here?
Thanks!
Related
I have a problem with drop_duplicates in a pandas dataframe. I'm importing lots of mixed data from an excel file into a dataframe and then doing various things to clean up the data. One of the stages is to remove any duplicates based on their coordinates.
In general this is working pretty well and importantly it's very fast, but I've had some problems and after an extensive search of the dataset I've found out that pandas is always a few duplicates.
Here's my test dataset:
x y z radius scale type
0 128.798699 76.038331 0.000 1.172 1.000 Node_B
1 136.373699 78.068331 0.000 1.172 1.000 Node_B
2 133.171699 74.866331 0.000 1.172 1.000 Node_B
3 135.201699 76.038331 0.000 1.172 1.000 Node_B
4 135.201699 82.442331 0.000 1.172 1.000 Node_B
5 136.373699 80.412331 0.000 1.172 1.000 Node_B
6 133.171699 83.614331 0.000 1.172 1.000 Node_B
7 127.626699 78.068331 0.000 1.172 1.000 Node_B
8 131.999699 79.240331 0.000 2.750 1.000 Node_A
9 90.199699 94.795331 0.626 0.325 0.650 Rib_B
10 85.799699 95.445331 0.626 0.325 0.650 Rib_B
11 90.199699 95.445331 0.626 0.325 0.650 Rib_B
12 91.865699 95.557331 0.537 0.438 0.876 Rib_B
13 128.798699 82.442331 0.000 1.172 1.000 Node_B
14 136.373699 80.412331 0.000 1.172 1.000 Node_B
15 158.373699 38.448331 0.000 1.172 1.000 Node_B
16 152.827699 35.246331 0.000 1.172 1.000 Node_B
17 157.201699 36.418331 0.000 1.172 1.000 Node_B
18 155.171699 35.246331 0.000 1.172 1.000 Node_B
19 215.626699 80.412331 0.000 1.172 1.000 Node_B
20 218.827699 83.614331 0.000 1.172 1.000 Node_B
21 216.798699 82.442331 0.000 1.172 1.000 Node_B
22 131.999699 79.240331 0.000 2.750 1.000 Node_A
23 128.798699 76.038331 0.000 1.172 1.000 Node_B
24 136.373699 78.068331 0.000 1.172 1.000 Node_B
25 162.051699 70.180331 0.626 0.325 0.650 Rib_D
26 162.619699 70.496331 0.626 0.325 0.650 Rib_D
27 189.948699 70.180331 0.626 0.325 0.650 Rib_D
I'm finding duplicates based on the x,y,z coordinates as these should be unique locations so I use df.drop_duplicates(subset=['x', 'y', 'z'], inplace=True) to remove any duplicates from the data frame. This seems to remove about 90% of my duplicates but it always seem to be missing some.
In the example dataframe there are several duplicates [0==23, 1==24, 6==14, 8==22] but pandas fails to remove them.
I found this using numpy and a very slow iterative loop that is comparing every point to every other point. It's ok for 50 or 100 points, but takes 15-20 minutes when I have 100-200K records in the dataframe.
How do I fix this? There is no precision parameter for drop_duplicates so why does it miss some?
You can use round as suggested by #mozway:
PRECISION = 3
df.drop(df[['x', 'y', 'z']].round(PRECISION).duplicated().loc[lambda x: x].index, inplace=True)
print(df)
# Output
x y z radius scale type
0 128.798699 76.038331 0.000 1.172 1.000 Node_B
1 136.373699 78.068331 0.000 1.172 1.000 Node_B
2 133.171699 74.866331 0.000 1.172 1.000 Node_B
3 135.201699 76.038331 0.000 1.172 1.000 Node_B
4 135.201699 82.442331 0.000 1.172 1.000 Node_B
5 136.373699 80.412331 0.000 1.172 1.000 Node_B
6 133.171699 83.614331 0.000 1.172 1.000 Node_B
7 127.626699 78.068331 0.000 1.172 1.000 Node_B
8 131.999699 79.240331 0.000 2.750 1.000 Node_A
9 90.199699 94.795331 0.626 0.325 0.650 Rib_B
10 85.799699 95.445331 0.626 0.325 0.650 Rib_B
11 90.199699 95.445331 0.626 0.325 0.650 Rib_B
12 91.865699 95.557331 0.537 0.438 0.876 Rib_B
13 128.798699 82.442331 0.000 1.172 1.000 Node_B
15 158.373699 38.448331 0.000 1.172 1.000 Node_B
16 152.827699 35.246331 0.000 1.172 1.000 Node_B
17 157.201699 36.418331 0.000 1.172 1.000 Node_B
18 155.171699 35.246331 0.000 1.172 1.000 Node_B
19 215.626699 80.412331 0.000 1.172 1.000 Node_B
20 218.827699 83.614331 0.000 1.172 1.000 Node_B
21 216.798699 82.442331 0.000 1.172 1.000 Node_B
25 162.051699 70.180331 0.626 0.325 0.650 Rib_D
26 162.619699 70.496331 0.626 0.325 0.650 Rib_D
27 189.948699 70.180331 0.626 0.325 0.650 Rib_D
trying to do polynomial regression and having some trouble fitting the model.
Getting
ValueError: Found input variables with inconsistent numbers of samples: [1040, 260]
import numpy as np
import pandas as pd
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures
x = BTCdata.iloc[:, [1, 2, 4, 5]]
y = BTCdata.iloc[:,3]
x, y = np.array(x).reshape((-1, 1)), np.array(y).reshape((-1, 1))
poly_features= PolynomialFeatures(degree= 4, include_bias = False)
x_ = poly_features.fit_transform(x)
model = LinearRegression()
model.fit(x_, y)
The problem comes from this line:
x = np.array(x).reshape((-1, 1))
By doing that you are transforming a dataframe of n rows and m columns into an array of n x m rows and 1 column. In your example, x ends up having 260 x 4 = 1040 rows whereas y has 260, raising this error.
If your goal is to convert your data to numpy arrays before using it in a model, then you can simply do:
x = x.to_numpy()
import numpy as np
import pandas as pd
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures
import statsmodels.api as sm
#
BTCdata = pd.read_excel('BitcoinRegression.xlsx', sheet_name='FinalBTC')
x = BTCdata.iloc[:, [1, 2, 4, 5]]
print(x.shape)
y = BTCdata.iloc[:,3]
print(y.shape)
#x, y = np.array(x).reshape((-1, 1)), np.array(y).reshape((-1, 1))
poly_features= PolynomialFeatures(degree= 4, include_bias = False)
x_ = poly_features.fit_transform(x)
#model = LinearRegression()
#model.fit(x_, y)
mod = sm.OLS(y, x_).fit()
mod.summary()
<class 'statsmodels.iolib.summary.Summary'>
"""
OLS Regression Results
==============================================================================
Dep. Variable: BTC R-squared: 0.886
Model: OLS Adj. R-squared: 0.868
Method: Least Squares F-statistic: 46.86
Date: Wed, 17 Mar 2021 Prob (F-statistic): 2.63e-85
Time: 20:49:58 Log-Likelihood: -2299.3
No. Observations: 260 AIC: 4675.
Df Residuals: 222 BIC: 4810.
Df Model: 37
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
x1 -0.0089 0.019 -0.468 0.640 -0.046 0.028
x2 0.0033 0.004 0.797 0.426 -0.005 0.012
x3 2.621e-05 3.55e-05 0.737 0.462 -4.38e-05 9.62e-05
x4 0.0005 0.001 0.789 0.431 -0.001 0.002
x5 -0.0238 0.067 -0.355 0.723 -0.156 0.108
x6 0.0790 0.688 0.115 0.909 -1.277 1.435
x7 0.0942 0.131 0.722 0.471 -0.163 0.352
x8 0.9679 1.276 0.759 0.449 -1.546 3.482
x9 0.0184 0.133 0.139 0.890 -0.243 0.280
x10 0.0093 0.013 0.726 0.469 -0.016 0.035
x11 0.0957 0.125 0.766 0.444 -0.150 0.342
x12 0.0001 0.000 0.864 0.389 -0.000 0.000
x13 0.0008 0.001 0.599 0.550 -0.002 0.003
x14 0.0207 0.026 0.783 0.435 -0.031 0.073
x15 3.594e-05 2.89e-05 1.245 0.214 -2.09e-05 9.28e-05
x16 -0.0004 0.001 -0.496 0.621 -0.002 0.001
x17 0.0158 0.010 1.621 0.106 -0.003 0.035
x18 -0.0068 0.002 -2.945 0.004 -0.011 -0.002
x19 -0.0014 0.007 -0.202 0.840 -0.015 0.012
x20 -0.0389 0.086 -0.454 0.650 -0.208 0.130
x21 0.1104 0.043 2.558 0.011 0.025 0.195
x22 0.7337 0.819 0.896 0.371 -0.881 2.348
x23 -1.4583 0.432 -3.378 0.001 -2.309 -0.607
x24 0.0601 0.031 1.913 0.057 -0.002 0.122
x25 0.0192 0.021 0.893 0.373 -0.023 0.061
x26 0.0403 0.091 0.445 0.657 -0.138 0.219
x27 -0.5110 0.224 -2.284 0.023 -0.952 -0.070
x28 0.0697 0.078 0.892 0.374 -0.084 0.224
x29 -0.1316 0.039 -3.397 0.001 -0.208 -0.055
x30 0.0054 0.103 0.052 0.958 -0.198 0.209
x31 0.0003 0.000 0.951 0.343 -0.000 0.001
x32 0.0060 0.007 0.856 0.393 -0.008 0.020
x33 -0.0124 0.012 -1.078 0.282 -0.035 0.010
x34 0.3317 0.394 0.842 0.400 -0.444 1.108
x35 -4.886e-09 1.1e-09 -4.439 0.000 -7.05e-09 -2.72e-09
x36 1.387e-07 3.68e-08 3.767 0.000 6.62e-08 2.11e-07
x37 5.106e-07 3.44e-06 0.148 0.882 -6.28e-06 7.3e-06
x38 4.652e-07 2.91e-07 1.601 0.111 -1.07e-07 1.04e-06
x39 -1.623e-06 5.17e-07 -3.138 0.002 -2.64e-06 -6.04e-07
x40 -8.446e-05 9.05e-05 -0.933 0.352 -0.000 9.39e-05
x41 -8.729e-06 7.38e-06 -1.182 0.238 -2.33e-05 5.82e-06
x42 -0.0017 0.002 -0.804 0.422 -0.006 0.002
x43 0.0007 0.000 1.705 0.090 -0.000 0.001
x44 -1.815e-05 2.11e-05 -0.862 0.390 -5.96e-05 2.33e-05
x45 9.562e-06 3.43e-06 2.788 0.006 2.8e-06 1.63e-05
x46 0.0012 0.001 1.413 0.159 -0.000 0.003
x47 5.405e-05 6.5e-05 0.831 0.407 -7.41e-05 0.000
x48 0.0069 0.044 0.156 0.876 -0.080 0.093
x49 -0.0078 0.006 -1.414 0.159 -0.019 0.003
x50 0.0001 0.000 0.307 0.759 -0.001 0.001
x51 0.1505 0.090 1.669 0.096 -0.027 0.328
x52 0.1555 0.046 3.410 0.001 0.066 0.245
x53 -0.0296 0.024 -1.210 0.227 -0.078 0.019
x54 0.0016 0.001 2.182 0.030 0.000 0.003
x55 -2.28e-05 8.77e-06 -2.600 0.010 -4.01e-05 -5.52e-06
x56 -0.0045 0.003 -1.594 0.112 -0.010 0.001
x57 -0.0002 0.000 -0.947 0.344 -0.001 0.000
x58 -0.0067 0.237 -0.028 0.977 -0.474 0.461
x59 0.0134 0.021 0.629 0.530 -0.029 0.055
x60 0.0020 0.002 1.123 0.262 -0.002 0.006
x61 0.0277 0.016 1.689 0.093 -0.005 0.060
x62 -0.3824 0.413 -0.926 0.355 -1.196 0.431
x63 0.3528 0.179 1.970 0.050 -0.000 0.706
x64 -0.0282 0.005 -5.708 0.000 -0.038 -0.018
x65 -0.0002 0.000 -0.695 0.488 -0.001 0.000
x66 0.0098 0.009 1.142 0.255 -0.007 0.027
x67 0.0901 0.103 0.873 0.384 -0.113 0.293
x68 -0.1941 0.648 -0.300 0.765 -1.471 1.083
x69 0.0237 0.021 1.128 0.261 -0.018 0.065
==============================================================================
Omnibus: 127.728 Durbin-Watson: 0.552
Prob(Omnibus): 0.000 Jarque-Bera (JB): 851.418
Skew: 1.861 Prob(JB): 1.31e-185
Kurtosis: 11.046 Cond. No. 4.00e+16
==============================================================================
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 4e+16. This might indicate that there are
strong multicollinearity or other numerical problems.
"""
I have dataframe, where i would like to replicate few rows:
X Y diff No
Index
1d 0.000 0.017 0.000e+00 0
2D 0.083 0.017 3.000e-03 1
3D 0.250 0.017 7.200e-03 2
6D 0.500 0.019 2.400e-03 3
1DD 1.000 0.020 2.400e-03 4
2DD 2.000 0.023 1.300e-03 5
3DD 3.000 0.024 1.000e-03 6
5DD 5.000 0.026 6.500e-04 7
7DD 7.000 0.027 2.667e-04 8
10DD 10.000 0.028 1.200e-04 9
20DD 20.000 0.029 1.200e-04 10
30DD 30.000 0.031 0.000e+00 11
I want to replicate 30DD 30 times and 20DD 20 times and 10DD 10 times with same index name.
I tried this, instead of replicating it multiplies
for i in range(4):
test1 = df.append(df.ix['30DD']*30)
X Y diff No
Index
1d 0.000 0.017 0.000e+00 0
2D 0.083 0.017 3.000e-03 1
3D 0.250 0.017 7.200e-03 2
6D 0.500 0.019 2.400e-03 3
1DD 1.000 0.020 2.400e-03 4
2DD 2.000 0.023 1.300e-03 5
3DD 3.000 0.024 1.000e-03 6
5DD 5.000 0.026 6.500e-04 7
7DD 7.000 0.027 2.667e-04 8
10DD 10.000 0.028 1.200e-04 9
20DD 20.000 0.029 1.200e-04 10
30DD 30.000 0.031 0.000e+00 11
30DD 900 0.918 0 330
Add new rows, but subtract 1, because append to original DataFrame:
vals = ['30DD'] * 29 + ['20DD'] * 19 + ['10DD'] * 9
df = df.append(df.loc[vals])
Last if want sorting values by numbers of index values:
df = df.iloc[df.index.str.extract('(\d+)').astype(int).squeeze().argsort()]
Using numpy.repeat, you can create a list of indices for rows you wish to append. Then feed to pd.DataFrame.loc and append to your original dataframe.
vals = ['30DD', '20DD', '10DD']
counts = [30, 20, 10]
df = df.append(df.loc[np.repeat(vals, counts)])
I am trying to create a rolling EWMA with the following decay= 1-ln(2)/3 on the last 13 values of a df such has :
factor
Out[36]:
EWMA
0 0.043
1 0.056
2 0.072
3 0.094
4 0.122
5 0.159
6 0.207
7 0.269
8 0.350
9 0.455
10 0.591
11 0.769
12 1.000
I have a df of monthly returns like this :
change.tail(5)
Out[41]:
date
2016-04-30 0.033 0.031 0.010 0.007 0.014 -0.006 -0.001 0.035 -0.004 0.020 0.011 0.003
2016-05-31 0.024 0.007 0.017 0.022 -0.012 0.034 0.019 0.001 0.006 0.032 -0.002 0.015
2016-06-30 -0.027 -0.004 -0.060 -0.057 -0.001 -0.096 -0.027 -0.096 -0.034 -0.024 0.044 0.001
2016-07-31 0.063 0.036 0.048 0.068 0.053 0.064 0.032 0.052 0.048 0.013 0.034 0.036
2016-08-31 -0.004 0.012 -0.005 0.009 0.028 0.005 -0.002 -0.003 -0.001 0.005 0.013 0.003
I am just trying to apply this rolling EWMA to each columns. I know that pandas has a EWMA method but I can't figure out how to pass the right 1-ln(2)/3 factor.
help would be appreciated! thanks!
#piRSquared 's answer is a good approximation, but values outside the last 13 also have weightings (albeit tiny), so it's not totally correct.
pandas could do rolling window calculations. However, amongst all the rolling function it supports, ewm is not one of them, which means we have to implement our own.
Assuming series is our time series to average:
from functools import partial
import numpy as np
window = 13
alpha = 1-np.log(2)/3 # This is ewma's decay factor.
weights = list(reversed([(1-alpha)**n for n in range(window)]))
ewma = partial(np.average, weights=weights)
rolling_average = series.rolling(window).apply(ewma)
use ewm with mean()
df.ewm(halflife=1 - np.log(2) / 3).mean()
I've got a pandas.DataFrame that looks like this:
>>> print df
0 1 2 3 4 5 6 7 8 9 10 11 \
0 0.198 0.198 0.266 0.198 0.236 0.199 0.198 0.198 0.199 0.199 0.199 0.198
1 0.032 0.034 0.039 0.405 0.442 0.382 0.343 0.311 0.282 0.255 0.232 0.210
2 0.702 0.702 0.742 0.709 0.755 0.708 0.708 0.712 0.707 0.706 0.706 0.706
3 0.109 0.112 0.114 0.114 0.128 0.532 0.149 0.118 0.115 0.114 0.114 0.112
4 0.309 0.306 0.311 0.311 0.316 0.513 1.977 0.313 0.311 0.310 0.311 0.309
5 0.280 0.277 0.282 0.278 0.282 0.383 1.122 1.685 0.280 0.280 0.282 0.280
6 0.466 0.460 0.465 0.465 0.468 0.508 0.829 1.100 1.987 0.465 0.465 0.463
7 0.469 0.464 0.469 0.470 0.469 0.490 0.648 0.783 1.095 2.002 0.469 0.466
8 0.137 0.120 0.137 0.138 0.137 0.136 0.144 0.149 0.166 0.209 0.137 0.136
9 0.125 0.107 0.125 0.126 0.125 0.122 0.126 0.128 0.132 0.144 0.125 0.123
10 0.125 0.106 0.125 0.123 0.123 0.122 0.125 0.128 0.132 0.142 0.125 0.123
11 0.127 0.107 0.125 0.125 0.125 0.122 0.126 0.127 0.132 0.142 0.125 0.123
12 0.125 0.107 0.125 0.128 0.125 0.123 0.126 0.127 0.132 0.142 0.125 0.122
13 0.871 0.862 0.871 0.872 0.872 0.872 0.873 0.872 0.875 0.880 0.873 0.872
14 0.114 0.115 0.116 0.117 0.131 0.536 0.153 0.123 0.118 0.117 0.117 0.116
15 0.033 0.032 0.031 0.032 0.032 0.040 0.033 0.033 0.032 0.032 0.032 0.032
12 13
0 0.198 0.198
1 0.190 0.172
2 0.705 0.705
3 0.112 0.115
4 0.308 0.310
5 0.275 0.278
6 0.462 0.463
7 0.466 0.466
8 0.134 1.678
9 0.122 1.692
10 0.122 1.694
11 0.122 1.695
12 0.122 1.684
13 0.872 1.255
14 0.116 0.127
15 0.031 0.032
[16 rows x 14 columns]
Each row represents a measurement value for an analog port. Each column is a test case. Thus there's one measurement for each of the analog ports, in each column.
When I plot this with DataFrame.plot() I end up with the following plot:
But this presents my rows, the 16 analog ports on the x-axis. I would like to have the column numbers on the x-axis. I've tried to define the x-axis in plot() as below:
>>> df.plot(x=df.columns)
Which results in a
ValueError: Length mismatch: Expected axis has 16 elements, new values have 14 elements
How should I approach this? Below is an example image which shows the correct x-axis values.
You want something like
df.T.plot()
Plus some other formatting. But that will get you started.
the .T method transposes the DataFrame.