Python

 

Intro

Le langage de programmation Python a été créé en 1989 par Guido van Rossum, aux Pays-Bas. Le nom Python vient d'un hommage à la série télévisée Monty Python's Flying Circus dont G. van Rossum est fan. La première version publique de ce langage a été publiée en 1991.

La Python Software Foundation est l'association qui organise le développement de Python et anime la communauté de développeurs et d'utilisateurs.

Caractéristiques:

      ·        Multiplateforme : fonctionnement multi-systèmes d'exploitation : Windows, Mac OS X, Linux, Android, iOS, depuis les mini-ordinateurs Raspberry Pi jusqu'aux supercalculateurs…

      ·        Gratuit : IDE/plateforme pour ordinateur, smartphone…

·         Langage de haut niveau : Besoin de quelques connaissances de base sur l’environnement informatique utilisé pour la programmation.
·         Langage interprété: Un script Python n'a pas besoin d'être compilé pour être exécuté, il est interprété en temps réel contrairement à des langages comme le C ou le C++.

·         Langage orienté objet : C'est-à-dire qu'il est possible de concevoir en Python des entités qui miment celles du monde réel (une cellule, une protéine, un atome, etc.) avec un certain nombre de règles de fonctionnement et d'interactions.
·         Domaine d’application fréquents : Analyse de données, robotique, IoT…

 

Introduction et shell

Le programme d'installation du Python est à partir de la page de téléchargements du site Python.org ( https://www.python.org/downloads/ )

Un shell est un interpréteur de commandes interactif permettant d'interagir avec l'ordinateur :

·         Shell Unix pour Mac OS X et Linux ;

·         Shell PowerShell pour Windows.

Un shell possède toujours une invite de commande ; Exemple pour le $ :

Lancer l’instruction « python » : Taper python après $ => $ python 

Exemple de lancement d’interrupteur de commandes python via Windows :

·         Power Shell

·         Invite commande:

Longue ligne de code

On peut la couper en deux avec le caractère \ (backslash) pour des raisons de lisibilité :

 

Commentaires

Le caractère # est utilisé pour insérer des commentaires ignorés par Python jusqu'à la fin de la ligne.

Indentation

Une indentation en Python doit être homogène (soit des espaces, soit des tabulations) pour un décalage d’un ensemble des lignes indentées constituant un bloc d'instructions.

Exécuter un programme

1.       Enregistrer (dans le dossier d’exécution de référence par défaut ; Exemple : C:\Users\aymen\ …) les instructions, appelé script ou programme, dans un fichier Python « nom_fichier.py ».

2.       Interpréter, via Shell (python test.py) ou interpréteur IDE…

Matlab serial

 How import data (Arduino...) from serial 

Xmess=[];

id_com='COM6';

%For safe connection

id = instrfind('port', id_com) ;

delete(id);

clear id;

%CR= detect end of line 

s = serial(id_com,'Terminator','CR');

fopen(s);

Xmess=sprintf('Serial State= %s Tranfert=%s',s.status,s.TransferStatus);

display(Xmess,'');

%testing serial buffer

isempty(line)

line=fgetl(s);

isempty(line)

while s.BytesAvailable>0

    %for  number reading from analog input: luminosity

    reading = fgetl(s);

    analog_lum=str2double(reading);

    

    %for  number reading from analog input: temperature

    reading=fscanf(s);

    val_tmp_c=str2double(reading);

    Xmess=sprintf('analog_Lum= %f  analog_tmp= %f',analog_lum,analog_tmp);

    display(Xmess,'');

    meas=fscanf(s);

    %isempty(line)

end

% Close and déconnection from serial port

fclose(s);

delete(s);

clear s;

Moderation of adolescents do hobbies within AIDS knowledge and current residence stability

Introduction

As said in the link,

The National Longitudinal Study of Adolescent Health (AddHealth) is a representative school-based survey of adolescents in grades 7-12 in the United States. The Wave 1 survey focuses on factors that may influence adolescents’ health and risk behaviors, including personal traits, families, friendships, romantic relationships, peer groups, schools, neighborhoods, and communities. 

source: Data Analysis Tools par Université Wesleyenne 

NB. This study is under Coursera training (Outils d'analyse des données) using python

Data

• H1GI3 (General Introductory)How old was the adolescent when he moved to his current residence?
  It's a quantitative value =>response (1, 2, 3 ... years)

• H1GH45 (General Health)

  How many people do you know who have AIDS?
  

  it's a quantitive value => response: range 0 to 98 people

• H1DA2 (Daily Activities)how many times did adolescents do hobbies? It's a categorical value => response ( not at all, 1 or 2 times, more, refused, don’t know...)

Objective

We study the moderation of did adolescents do hobbies about the current residence period and AIDs existence in neighborhood knowledge.

• N.B As explanatory and response used data are both quantitive, we would like to use Pearson correlation (r correlation analysis).

Python code

# -*- coding: utf-8 -*-"""
Created on Mon Apr 20 00:30:09 2020
@author: ABID Aymen
"""
# CORRELATION
import pandas
import numpy
import scipy.stats
import seaborn
import matplotlib.pyplot as plt
mydata = pandas.read_csv('addhealth_pds.csv', low_memory=False)
mydata['H1GI3'] = mydata['H1GI3'].apply(pandas.to_numeric, errors='coerce')
mydata['H1GH45'] = mydata['H1GH45'].apply(pandas.to_numeric, errors='coerce')
mydata['H1DA2'] = mydata['H1DA2'].apply(pandas.to_numeric, errors='coerce')
#data cleaning from no input values
mydata['H1DA2']=mydata['H1DA2'].replace(' ', numpy.nan)
data_clean=mydata.dropna()
#%%
#pearson information (r,p)
print ('Association between residence stability and AIDS people knowledge  (r,p)')
[r,p]=scipy.stats.pearsonr(data_clean['H1GI3'], data_clean['H1GH45'])
print([r,p])
#moderator grouping
def incomegrp (row):
   if row['H1DA2'] <= 3:
      return 1 
   elif row['H1DA2'] > 6:
      return 2
data_clean['incomegrp'] = data_clean.apply (lambda row: incomegrp (row),axis=1)
chk1 = data_clean['incomegrp'].value_counts(sort=False, dropna=False)
print(chk1)
sub1=data_clean[(data_clean['incomegrp']== 1)]
sub2=data_clean[(data_clean['incomegrp']== 2)]
#pearson information (r,p) per subgroup (sub1..)
print ('Association between uresidence stability and AIDS people knowledge for exact answer')
print (scipy.stats.pearsonr(sub1['H1GI3'], sub1['H1GH45']))
print ('       ')
print ('Association between uresidence stability and AIDS people knowledge for not exact answer')
print (scipy.stats.pearsonr(sub2['H1GI3'], sub2['H1GH45']))
#%%
#graphics
scat1 = seaborn.regplot(x="H1GI3", y="H1GH45", data=sub1)
plt.xlabel('Residence Rate')
plt.ylabel('AIDS people knowledge Rate')
plt.title('Scatterplot for the Association Between Urban Rate and Internet Use Rate for exact answer countries')
print (scat1)
#%%
scat2 = seaborn.regplot(x="H1GI3", y="H1GH45", fit_reg=False, data=sub2)
plt.xlabel('Residence Rate')
plt.ylabel('AIDS people knowledge Rate')
plt.title('Scatterplot for the Association Between Urban Rate and Internet Use Rate for not exact answer')
print (scat2)

Results

Association between residence stability and AIDS people knowledge  (r,p)

[0.15162579814828608, 9.506977908603123e-35]

Association between uresidence stability and AIDS people knowledge for exact answer

(0.034754024175195075, 0.005088079230752868)       

Association between uresidence stability and AIDS people knowledge for not exact answer

(1.0, 0.0)

So,

For the exact answer income group the correlation between 

residence stability and AIDS people knowledge are 0.15 and the P-value is so significant. 

 

For the not exact answer, the association between 

 

residence stability and AIDS people knowledge is 0.034 less

 

 significant P-value

 

.

 

When we map these findings onto the associated scatter plots for 

 

each income group, we are better able to visualize the significant and 

 

non-significant relationships. 

 

Estimating a line of best fit within each scatter plot 

 

shows the positive association between 

 

residence stability and AIDS people knowledge use rate among the exact responses more than not exact (orange points).

Correlation for AIDS knowledge using current residence stability

Introduction

As said in the link,

The National Longitudinal Study of Adolescent Health (AddHealth) is a representative school-based survey of adolescents in grades 7-12 in the United States. The Wave 1 survey focuses on factors that may influence adolescents’ health and risk behaviors, including personal traits, families, friendships, romantic relationships, peer groups, schools, neighborhoods, and communities.

source: Data Analysis Tools par Université Wesleyenne 

NB. This study is under Coursera training (Outils d'analyse des données) using python

Data

• H1GI3 (General Introductory)How old was the adolescent when he moved to his current residence?
  It's a quantitative value =>response (1, 2, 3 ... years)

• H1GH45 (General Health)
  

  How many people do you know who have AIDS?
  

  it's a quantitive value => response: range 0 to 98 people

Objective:

Study of the correlation between time spent by adolescents in current residence and AIDS people in the neighborhood.

Python code:

# -*- coding: utf-8 -*-

"""

Created on Sun Apr 19 17:51:51 2020

@author: Aymen ABID

"""

import pandas

import numpy

import seaborn

import scipy

import matplotlib.pyplot as plt

import math  

mydata = pandas.read_csv('addhealth_pds.csv', low_memory=False)

mydata['H1GI3'] = mydata['H1GI3'].apply(pandas.to_numeric, errors='coerce')

mydata['H1GH45'] = mydata['H1GH45'].apply(pandas.to_numeric, errors='coerce')

mydata['H1GI3'] = mydata['H1GI3'].apply(pandas.to_numeric, errors='coerce')

mydata['H1GH45'] = mydata['H1GH45'].apply(pandas.to_numeric, errors='coerce')

mydata['H1GI3']=mydata['H1GI3'].replace(' ', numpy.nan)

mydata['H1GH45']=mydata['H1GH45'].replace(' ', numpy.nan)

data_clean=mydata.dropna() #clean data from Nan measurements for correlation

print ('Association between residence stability and AIDS people knowledge  (r,p)')

[r,p]=scipy.stats.pearsonr(data_clean['H1GI3'], data_clean['H1GH45'])

print([r,p])

print('Fraction of the variability ')

print (math.sqrt( r))

scat1 = seaborn.regplot(x="H1GI3", y="H1GH45", fit_reg=True, data=mydata)

plt.xlabel('Residence Rate')

plt.ylabel('AIDS people knowledge Rate')

plt.title('Association Between time in current residence and AIDS that adolescent know')

Results:

Association between residence stability and AIDS people knowledge  (r,p)

[0.15162579814828608, 9.506977908603123e-35]

Fraction of the variability 

0.3893915743159912

So,

The association between urbanrate and internetuserate, the correlation 

coefficient is approximately 0.15, with a very small p-value.

The association between current residence period rate and AIDS pepoplle in neghborhood knowlge is fairly strong, 

and it's also positive as the scatter plot has already shown us. 

The association between two analysed rates is also positive

.

If we square our correlation coefficient of 0.15, 

we get 0.38. 

This could be interpreted the following way. 

If we know the residence rate, 

we can predict 38% of the variability we will see in the rate of AIDS knowlage. 

Of course, that also means that 62% of the variability is unaccounted for it. 

Chi-Square Test: thinking about the home of childhood and daily activities such hobbies

Introduction

As said in link,

The National Longitudinal Study of Adolescent Health (AddHealth) is a representative school-based survey of adolescents in grades 7-12 in the United States. The Wave 1 survey focuses on factors that may influence adolescents’ health and risk behaviors, including personal traits, families, friendships, romantic relationships, peer groups, schools, neighborhoods, and communities.

source: Data Analysis Tools par Université Wesleyenne 

NB. This study is under Coursera training (Outils d'analyse des données) using python

Data

H1GI2 (General Introductory) : 

The question was:
Think about the house or apartment building in which you lived in January 1990, when you were {AGE IN JANUARY 1990} years old. Do you still live there? 

Frequency	Code	Response
3046	0	no
3447	1	yes
3	6	refused
8	8	don't know

=> 4 categories

H1DA2 (Daily Activities): 

The question was:

How many times did adolescents do hobbies? It's a categorical value => response ( not at all, 1 or 2 times, more, refused, don’t know...)

Frequency	Code	Response
1416	0	not at all
2163	1	1 or 2 times
1439	2	3 or 4 times
1479	3	5 or more times
2	6	refused
5	8	don't know

=> 6 catégories

Objective:

Study of the impact & relationship between thinking about the home of childhood and daily activities such hobbies.

code:

# -*- coding: utf-8 -*-
"""
Created on Sat Apr 18 13:11:16 2020

@author: PC HP
"""
import pandas
import numpy
import scipy.stats
import seaborn
import matplotlib.pyplot as plt
mydata = pandas.read_csv('addhealth_pds.csv', low_memory=False)
mydata['H1GI2'] = mydata['H1GI2'].apply(pandas.to_numeric, errors='coerce')
mydata['H1DA2'] = mydata['H1DA2'].apply(pandas.to_numeric, errors='coerce')
#mydata['H1GI2'] = pandas.to_numeric(mydata['H1GI2'], errors='coerce')
#mydata['H1DA2 '] = pandas.to_numeric(mydata['H1DA2 '], errors='coerce')
#SETTING MISSING DATA
mydata['H1GI2']=mydata['H1GI2'].replace(200, numpy.nan)
mydata['H1DA2']=mydata['H1DA2'].replace(200, numpy.nan)
# contingency table of observed counts
print('observed counts table:')
ct1=pandas.crosstab(mydata['H1GI2'], mydata['H1DA2'])
print (ct1)
# column percentages
colsum=ct1.sum(axis=0)
colpct=ct1/colsum
print('chi percentages table:')
print(colpct)
# chi-square
print ('chi-square value, p value, expected counts')
cs1= scipy.stats.chi2_contingency(ct1)
print (cs1)
# set variable types 
mydata["H1DA2"] = mydata["H1DA2"].astype('category')
# new code for setting variables to numeric:
mydata['H1GI2'] = pandas.to_numeric(mydata['H1GI2'], errors='coerce')
# graph
seaborn.catplot(x="H1DA2", y="H1GI2", data=mydata, kind="bar", ci=None)
plt.xlabel('Times did adolescents do hobbies last week')
plt.ylabel('Proportion residence leaving in childhood Dependent')

recode = {0: 0}
mydata['subH1DA2']= mydata['H1DA2'].map(recode)
# contingency table of observed counts
ct2=pandas.crosstab(mydata['H1GI2'], mydata['subH1DA2'])
print (ct2)
# column percentages
colsum=ct2.sum(axis=0)
colpct=ct2/colsum
print(colpct)
print ('chi-square value, p value, expected counts')
cs2= scipy.stats.chi2_contingency(ct2)
print (cs2)
#next sub groups
recode = {0: 0,1:1}
mydata['subH1DA2']= mydata['H1DA2'].map(recode)
# contingency table of observed counts
ct2=pandas.crosstab(mydata['H1GI2'], mydata['subH1DA2'])
print (ct2)
# column percentages
colsum=ct2.sum(axis=0)
colpct=ct2/colsum
print(colpct)
print ('chi-square value, p value, expected counts')
cs2= scipy.stats.chi2_contingency(ct2)
print (cs2)
#next sub groups
recode = {0: 0,1:1,2:2}
mydata['subH1DA2']= mydata['H1DA2'].map(recode)
# contingency table of observed counts
ct2=pandas.crosstab(mydata['H1GI2'], mydata['subH1DA2'])
print (ct2)
# column percentages
colsum=ct2.sum(axis=0)
colpct=ct2/colsum
print(colpct)
print ('chi-square value, p value, expected counts')
cs2= scipy.stats.chi2_contingency(ct2)
print (cs2)
#next sub groups
recode = {0: 0,1:1,2:2,3:3}
mydata['subH1DA2']= mydata['H1DA2'].map(recode)
# contingency table of observed counts
ct2=pandas.crosstab(mydata['H1GI2'], mydata['subH1DA2'])
print (ct2)
# column percentages
colsum=ct2.sum(axis=0)
colpct=ct2/colsum
print(colpct)
print ('chi-square value, p value, expected counts')
cs2= scipy.stats.chi2_contingency(ct2)
print (cs2)
#next sub groups
recode = {0: 0,1:1,2:2,3:3,6:6}
mydata['subH1DA2']= mydata['H1DA2'].map(recode)
# contingency table of observed counts
ct2=pandas.crosstab(mydata['H1GI2'], mydata['subH1DA2'])
print (ct2)
# column percentages
colsum=ct2.sum(axis=0)
colpct=ct2/colsum
print(colpct)
print ('chi-square value, p value, expected counts')
cs2= scipy.stats.chi2_contingency(ct2)
print (cs2)
#next sub groups
recode = {0: 0,1:1,2:2,3:3,8:8}
mydata['subH1DA2']= mydata['H1DA2'].map(recode)
# contingency table of observed counts
ct2=pandas.crosstab(mydata['H1GI2'], mydata['subH1DA2'])
print (ct2)
# column percentages
colsum=ct2.sum(axis=0)
colpct=ct2/colsum
print(colpct)
print ('chi-square value, p value, expected counts')
cs2= scipy.stats.chi2_contingency(ct2)
print (cs2)
#next sub groups
recode = {6:6,8:8}
mydata['subH1DA2']= mydata['H1DA2'].map(recode)
# contingency table of observed counts
ct2=pandas.crosstab(mydata['H1GI2'], mydata['subH1DA2'])
print (ct2)
# column percentages
colsum=ct2.sum(axis=0)
colpct=ct2/colsum
print(colpct)
print ('chi-square value, p value, expected counts')
cs2= scipy.stats.chi2_contingency(ct2)
print (cs2)

Results:

According to the figure the S1 = {x = [0..3]} are in relation and for S2 = {x = [6.8]} with a low probability. But the two subgroups are not dependant. This is explained also by the following:

• S1:

subH1DA2       0.0       1.0       2.0       3.0 H1GI2                                           0         0.496469  0.478502  0.451703  0.444219 1         0.500706  0.521036  0.548297  0.555105 6         0.000706  0.000000  0.000000  0.000000 8         0.002119  0.000462  0.000000  0.000676 chi-square value, p value, expected counts (19.160401671234695, 0.023863055784254898, 9, array([[6.63647837e+02, 1.01375019e+03, 6.74427428e+02, 6.93174542e+02],        [7.51044482e+02, 1.14725227e+03, 7.63243651e+02, 7.84459597e+02],        [2.17946745e-01, 3.32922887e-01, 2.21486840e-01, 2.27643528e-01],        [1.08973372e+00, 1.66461444e+00, 1.10743420e+00, 1.13821764e+00]]))

• S2:

subH1DA2  6.0  8.0 H1GI2             0           0    1 1           0    1 6           2    0 8           0    3 subH1DA2  6.0  8.0 H1GI2             0         0.0  0.2 1         0.0  0.2 6         1.0  0.0 8         0.0  0.6 chi-square value, p value, expected counts (7.0, 0.07189777249646509, 3, array([[0.28571429, 0.71428571],        [0.28571429, 0.71428571],        [0.57142857, 1.42857143],        [0.85714286, 2.14285714]]))

Indeed, the value of chi and p gave for S1 an important chi even if p weak which means the negation of H0. Whereas for S2, a smaller chi and a higher p than S1, which reduces to the dependence with the less negotiation for H0.
More details are available in the following results (including post hoc analysis):

observed counts table: H1DA2    0     1    2    3  6  8 H1GI2                           0      703  1035  650  657  0  1 1      709  1127  789  821  0  1 6        1     0    0    0  2  0 8        3     1    0    1  0  3 chi percentages table: H1DA2         0         1         2         3    6    8 H1GI2                                                  0      0.496469  0.478502  0.451703  0.444219  0.0  0.2 1      0.500706  0.521036  0.548297  0.555105  0.0  0.2 6      0.000706  0.000000  0.000000  0.000000  1.0  0.0 8      0.002119  0.000462  0.000000  0.000676  0.0  0.6 chi-square value, p value, expected counts (5810.662666395812, 0.0, 15, array([[6.63151292e+02, 1.01299170e+03, 6.73922817e+02, 6.92655904e+02,         9.36654367e-01, 2.34163592e+00],        [7.50453875e+02, 1.14635009e+03, 7.62643450e+02, 7.83842712e+02,         1.05996310e+00, 2.64990775e+00],        [6.53136531e-01, 9.97693727e-01, 6.63745387e-01, 6.82195572e-01,         9.22509225e-04, 2.30627306e-03],        [1.74169742e+00, 2.66051661e+00, 1.76998770e+00, 1.81918819e+00,         2.46002460e-03, 6.15006150e-03]])) subH1DA2  0.0 H1GI2        0         703 1         709 6           1 8           3 subH1DA2       0.0 H1GI2             0         0.496469 1         0.500706 6         0.000706 8         0.002119 chi-square value, p value, expected counts (0.0, 1.0, 0, array([[703.],        [709.],        [  1.],        [  3.]])) subH1DA2  0.0   1.0 H1GI2              0         703  1035 1         709  1127 6           1     0 8           3     1 subH1DA2       0.0       1.0 H1GI2                       0         0.496469  0.478502 1         0.500706  0.521036 6         0.000706  0.000000 8         0.002119  0.000462 chi-square value, p value, expected counts (4.886483669772731, 0.18030059764588033, 3, array([[6.87624476e+02, 1.05037552e+03],        [7.26397318e+02, 1.10960268e+03],        [3.95641241e-01, 6.04358759e-01],        [1.58256496e+00, 2.41743504e+00]])) subH1DA2  0.0   1.0  2.0 H1GI2                   0         703  1035  650 1         709  1127  789 6           1     0    0 8           3     1    0 subH1DA2       0.0       1.0       2.0 H1GI2                                 0         0.496469  0.478502  0.451703 1         0.500706  0.521036  0.548297 6         0.000706  0.000000  0.000000 8         0.002119  0.000462  0.000000 chi-square value, p value, expected counts (13.279010760752726, 0.03881309634250806, 6, array([[6.73855719e+02, 1.02934316e+03, 6.84801116e+02],        [7.40733360e+02, 1.13150159e+03, 7.52765046e+02],        [2.82184137e-01, 4.31048226e-01, 2.86767637e-01],        [1.12873655e+00, 1.72419291e+00, 1.14707055e+00]])) subH1DA2  0.0   1.0  2.0  3.0 H1GI2                        0         703  1035  650  657 1         709  1127  789  821 6           1     0    0    0 8           3     1    0    1 subH1DA2       0.0       1.0       2.0       3.0 H1GI2                                            0         0.496469  0.478502  0.451703  0.444219 1         0.500706  0.521036  0.548297  0.555105 6         0.000706  0.000000  0.000000  0.000000 8         0.002119  0.000462  0.000000  0.000676 chi-square value, p value, expected counts (19.160401671234695, 0.023863055784254898, 9, array([[6.63647837e+02, 1.01375019e+03, 6.74427428e+02, 6.93174542e+02],        [7.51044482e+02, 1.14725227e+03, 7.63243651e+02, 7.84459597e+02],        [2.17946745e-01, 3.32922887e-01, 2.21486840e-01, 2.27643528e-01],        [1.08973372e+00, 1.66461444e+00, 1.10743420e+00, 1.13821764e+00]])) subH1DA2  0.0   1.0  2.0  3.0  6.0 H1GI2                             0         703  1035  650  657    0 1         709  1127  789  821    0 6           1     0    0    0    2 8           3     1    0    1    0 subH1DA2       0.0       1.0       2.0       3.0  6.0 H1GI2                                                0         0.496469  0.478502  0.451703  0.444219  0.0 1         0.500706  0.521036  0.548297  0.555105  0.0 6         0.000706  0.000000  0.000000  0.000000  1.0 8         0.002119  0.000462  0.000000  0.000676  0.0 chi-square value, p value, expected counts (4348.773173724676, 0.0, 12, array([[6.63443607e+02, 1.01343822e+03, 6.74219880e+02, 6.92961225e+02,         9.37067241e-01],        [7.50813356e+02, 1.14689922e+03, 7.63008771e+02, 7.84218187e+02,         1.06047084e+00],        [6.53639021e-01, 9.98461302e-01, 6.64256039e-01, 6.82720419e-01,         9.23218957e-04],        [1.08939837e+00, 1.66410217e+00, 1.10709340e+00, 1.13786736e+00,         1.53869826e-03]])) subH1DA2  0.0   1.0  2.0  3.0  8.0 H1GI2                             0         703  1035  650  657    1 1         709  1127  789  821    1 6           1     0    0    0    0 8           3     1    0    1    3 subH1DA2       0.0       1.0       2.0       3.0  8.0 H1GI2                                                0         0.496469  0.478502  0.451703  0.444219  0.2 1         0.500706  0.521036  0.548297  0.555105  0.2 6         0.000706  0.000000  0.000000  0.000000  0.0 8         0.002119  0.000462  0.000000  0.000676  0.6 chi-square value, p value, expected counts (1477.2704083643332, 3.0249713664075e-309, 12, array([[6.63355275e+02, 1.01330329e+03, 6.74130114e+02, 6.92868963e+02,         2.34235620e+00],        [7.50684712e+02, 1.14670271e+03, 7.62878038e+02, 7.84083820e+02,         2.65072285e+00],        [2.17779145e-01, 3.32666872e-01, 2.21316518e-01, 2.27468471e-01,         7.68994156e-04],        [1.74223316e+00, 2.66133497e+00, 1.77053214e+00, 1.81974777e+00,         6.15195325e-03]])) subH1DA2  6.0  8.0 H1GI2             0           0    1 1           0    1 6           2    0 8           0    3 subH1DA2  6.0  8.0 H1GI2             0         0.0  0.2 1         0.0  0.2 6         1.0  0.0 8         0.0  0.6 chi-square value, p value, expected counts (7.0, 0.07189777249646509, 3, array([[0.28571429, 0.71428571],        [0.28571429, 0.71428571],        [0.57142857, 1.42857143],        [0.85714286, 2.14285714]]))

Accroding to just above resultes, we abserve that if data of 'x' (subH1DA2 from H1DA2) is not contains x in {6,8}  we have good independance with poor p. In adding 6 or 8 data or all, the chi value is more important and p converge to zero.

So, we conclude the rejection of H0 with great independence between thinking about the home of childhood and daily activities such as hobbies. Also, according to post hoc comparison, it exists great independence between response subgroups; who would like to give response x= [0ènot at all..., 3è5 or more time] or who don't give (refused, don' know) according to their recent hobbies doing.