.. _2018-09-18rappelspythonrst:

==========================================
2018-09-18 - Rappels sur le langage python
==========================================


.. only:: html

    **Links:** :download:`notebook <2018-09-18_rappels_python.ipynb>`, :downloadlink:`html <2018-09-18_rappels_python2html.html>`, :download:`python <2018-09-18_rappels_python.py>`, :downloadlink:`slides <2018-09-18_rappels_python.slides.html>`, :githublink:`GitHub|_doc/notebooks/notebook_eleves/2018-2019/2018-09-18_rappels_python.ipynb|*`


Notebooks écrit durant la séance du 18 septembre 2018 à propos du
langage Python.

.. code:: ipython3

    a = 4

.. code:: ipython3

    a




.. parsed-literal::
    4



.. code:: ipython3

    print(a)


.. parsed-literal::
    4


.. code:: ipython3

    b = 5
    c = a + b
    c




.. parsed-literal::
    9



.. code:: ipython3

    entier = 1
    reel = 4.5
    chaine = "45"
    chaine = '45'
    chaine = """
    fgfgqlibufliszqb
    """

.. code:: ipython3

    chaine




.. parsed-literal::
    '\nfgfgqlibufliszqb\n'



.. code:: ipython3

    print(chaine)


.. parsed-literal::

    fgfgqlibufliszqb
    


.. code:: ipython3

    couple = (4, 5)
    couple




.. parsed-literal::
    (4, 5)



.. code:: ipython3

    nuplet = (4, 6, 7, 'e')

.. code:: ipython3

    nuplet




.. parsed-literal::
    (4, 6, 7, 'e')



.. code:: ipython3

    liste = [4, 6, 7, 'e']
    liste




.. parsed-literal::
    [4, 6, 7, 'e']



.. code:: ipython3

    nuplet[1]




.. parsed-literal::
    6



.. code:: ipython3

    liste[1]




.. parsed-literal::
    6



.. code:: ipython3

    nuplet[1] = 5


::

    ---------------------------------------------------------------------------
    TypeError                                 Traceback (most recent call last)
    <ipython-input-18-5af74e621b46> in <module>()
    ----> 1 nuplet[1] = 5
    
    TypeError: 'tuple' object does not support item assignment


.. code:: ipython3

    liste[1] = 5
    liste




.. parsed-literal::
    [4, 5, 7, 'e']



.. code:: ipython3

    dico = {'e': 4}
    dico




.. parsed-literal::
    {'e': 4}



.. code:: ipython3

    dico['e']




.. parsed-literal::
    4



.. code:: ipython3

    dico[0]  ##


::

    ---------------------------------------------------------------------------
    KeyError                                  Traceback (most recent call last)
    <ipython-input-22-08669366fa89> in <module>()
    ----> 1 dico[0]
    
    KeyError: 0


.. code:: ipython3

    dico[0] = 5
    dico




.. parsed-literal::
    {'e': 4, 0: 5}



.. code:: ipython3

    dico['lirbfvvbfsjq'] = 'lfvlkjqslkjdf'

.. code:: ipython3

    dico




.. parsed-literal::
    {'e': 4, 0: 5, 'lirbfvvbfsjq': 'lfvlkjqslkjdf'}



.. code:: ipython3

    dico[(4,5)] = 5

.. code:: ipython3

    dico




.. parsed-literal::
    {'e': 4, 0: 5, 'lirbfvvbfsjq': 'lfvlkjqslkjdf', (4, 5): 5}



.. code:: ipython3

    dico[ [4,6] ] = 6


::

    ---------------------------------------------------------------------------
    TypeError                                 Traceback (most recent call last)
    <ipython-input-28-3097bf2b8dc7> in <module>()
    ----> 1 dico[ [4,6] ] = 6
    
    TypeError: unhashable type: 'list'


.. code:: ipython3

    texte = "llmmaab"
    dico = {}
    for c in texte:
        if c in dico:
            print('+1')
            dico[c] = dico[c] + 1
        elif c == 'l':
            print('+2')
            dico[c] = 0.5
        else:
            print('+3')
            dico[c] = 1
        print(c, dico)
    dico


.. parsed-literal::
    +2
    l {'l': 0.5}
    +1
    l {'l': 1.5}
    +3
    m {'l': 1.5, 'm': 1}
    +1
    m {'l': 1.5, 'm': 2}
    +3
    a {'l': 1.5, 'm': 2, 'a': 1}
    +1
    a {'l': 1.5, 'm': 2, 'a': 2}
    +3
    b {'l': 1.5, 'm': 2, 'a': 2, 'b': 1}




.. parsed-literal::

    {'l': 1.5, 'm': 2, 'a': 2, 'b': 1}



.. code:: ipython3

    def fonction_histogramme(texte):
        dico = {}
        for c in texte:
            if c in dico:
                dico[c] = dico[c] + 1
            elif c == 'l':
                dico[c] = 0.5
            else:
                dico[c] = 1
        return dico
    
    fonction_histogramme('eriuhgier')




.. parsed-literal::
    {'e': 2, 'r': 2, 'i': 2, 'u': 1, 'h': 1, 'g': 1}



.. code:: ipython3

    dico = fonction_histogramme('eriuhgier')
    print(dico['e'])


.. parsed-literal::
    2