For a list of learning Python dictionary

  List

  Python list is an ordered collection of the most flexible type of object.

  # List iteration and resolve

  >>> res = [c*4 for c in 'Spam']

  >>> res

  ['SSSS', 'pppp', 'aaaa', 'mmmm'

  >>> res = []

  >>> for c in 'Spam':

  ... res.append(c*4)

  ...

  >>> res

  ['SSSS', 'pppp', 'aaaa', 'mmmm']

  >>> list(map(abs,[-1,-2,0,1,2]))

  [1, 2, 0, 1, 2]

  # General Operation

  >>> L = [5,3,6,2,8]

  >>> sorted(L)

  [2, 3, 5, 6, 8]

  >>> L

  [5, 3, 6, 2, 8]

  >>> L.sort()

  >>> L

  [2, 3, 5, 6, 8]

  >>> L.insert(0,1)

  >>> L

  [1, 2, 3, 5, 6, 8]

  >>> L.reverse()

  >>> L

  [8, 6, 5, 3, 2, 1]

  >>>

  >>> del L[0]

  >>> L

  [6, 5, 3, 2, 1]

  >>> L.pop()

  1

  >>> L

  [6, 5, 3, 2]

  >>> L.remove(6)

  >>> L

  [5, 3, 2]

  '''

  Place the list of modifications: Because Python only deal with object references, so you need to modify the original objects at a separate area with a new object generated.

  Because when you modify an object in place, may also affect more than one reference to it.

  '''

  Other #

  >>> L = ['already','got','one']

  >>> L

  ['already', 'got', 'one']

  >>> L[1:]=[]

  >>> L

  ['already']

  >>> L[0]=[]

  >>> L

  [[]]

  dictionary

  If the list is ordered as a collection of objects, dictionaries can be treated as unordered collection. The main difference is that: among the elements of the dictionary is accessed by a key, rather than by shifting access.

  >>> D = {'food':{'ham':1,'egg':2}}

  >>> D.get('food')

  { 'Him': 1, 'egg': 2}

  >>> D2 = {'a':1,'b':2}

  >>> D.update(D2)

  >>> D

  {'food': {'ham': 1, 'egg': 2}, 'a': 1, 'b': 2}

  >>> D.pop('b')

  2

  >>> len(D)

  2

  >>> D

  {'food': {'ham': 1, 'egg': 2}, 'a': 1}

  >>> del D['a']

  >>> D

  {'food': {'ham': 1, 'egg': 2}}

  >>> D = {x:x*2 for x in range(10)}

  >>> D

  {0: 0, 1: 2, 2: 4, 3: 6, 4: 8, 5: 10, 6: 12, 7: 14, 8: 16, 9: 18}

  >>> D = {'spam':2,'ham':1,'eggs':3}

  >>> D['spam']

  2

  >>> D

  {'spam': 2, 'ham': 1, 'eggs': 3}

  >>> len(D)

  3

  >>> 'ham' in D

  True

  >>> list(D.keys())

  ['spam', 'ham', 'eggs']

  >>> list(D.values())

  [2, 1, 3]

  >>> D

  {'spam': 2, 'ham': 1, 'eggs': 3}

  >>> D['ham']=['grill','bake','fry']

  >>> D Wuxi look good Men's Hospital Which https://yyk.familydoctor.com.cn/20612/

  {'spam': 2, 'ham': ['grill', 'bake', 'fry'], 'eggs': 3}

  >>> del D['eggs']

  >>> D

  {'spam': 2, 'ham': ['grill', 'bake', 'fry']}

  >>> D['brunch'] = 'Bacon'

  >>> D

  {'spam': 2, 'ham': ['grill', 'bake', 'fry'], 'brunch': 'Bacon'}

  >>> list(D.items())

  [('spam', 2), ('ham', ['grill', 'bake', 'fry']), ('brunch', 'Bacon')]

  >>> D2 = {'toast':4,'muffin':5}

  >>> D.update(D2)

  >>> D

  {'spam': 2, 'ham': ['grill', 'bake', 'fry'], 'brunch': 'Bacon', 'toast': 4, 'muffin': 5}

  >>> table = {'Python':'Guido van Rossum',

  ... 'Perl':'Larry Wall',

  ... 'Tcl':'John Ousterhout'}

  >>>

  >>> language = 'Python'

  >>> creator = table[language]

  >>> creator

  "Guido van Rossum '

  >>> for lang in table:

  ... print (just, '\ t', table [only])

  ...

  Python Guido van Rossum

  Perl Larry Wall

  Tcl John Ousterhout

  # Three ways to avoid missing-key error

  ... if (2,3,6) in Matrix:

  ... print(Matrix[(2,3,6)])

  ... else:

  ... print(0)

  ...

  0

  >>>

  >>>

  >>> try:

  ... print(Matrix[(2,3,6)])

  ... except KeyError:

  ... print(0)

  ...

  0

  >>>

  >>>

  >>> Matrix.get((2,3,4),0)

  98

  >>> Matrix.get((2,3,6),0)

  0

Guess you like

Origin www.cnblogs.com/djw12333/p/11264077.html