一本简单的字典:

x = {'X':"yes", 'Y':"no", 'Z':"ok"}

在Python 3中打印一个特定的(键，值)对(本例中是索引1处的pair):

for e in range(len(x)):
    print(([x for x in x.keys()][e], [x for x in x.values()][e]))

输出:

('X', 'yes')
('Y', 'no')
('Z', 'ok')

下面是打印元组中所有对的一行程序:

print(tuple(([x for x in x.keys()][i], [x for x in x.values()][i]) for i in range(len(x))))

输出:

(('X', 'yes'), ('Y', 'no'), ('Z', 'ok'))

Python 2和Python 3

I是键，所以你只需要使用它:

for i in d:
    print i, d[i]

Python 3

D.items()返回迭代器;要获得一个列表，需要自己将迭代器传递给list()。

for k, v in d.items():
    print(k, v)

Python 2

你可以得到一个同时包含键和值的迭代器。D.items()返回一个(key, value)元组列表，而d.iteritems()返回一个提供相同元组的迭代器:

for k, v in d.iteritems():
    print k, v

for key, value in d.iteritems():
    print key, '\t', value

对于Python 3.x

for key, value in d.iteritems():
    print(f'{key}\t{value}') 

除了已经提到的方法之外。可以使用" viewitems " " viewkeys " " viewvalues "

>>> d = {320: 1, 321: 0, 322: 3}
>>> list(d.viewitems())
[(320, 1), (321, 0), (322, 3)]
>>> list(d.viewkeys())
[320, 321, 322]
>>> list(d.viewvalues())
[1, 0, 3]

Or

>>> list(d.iteritems())
[(320, 1), (321, 0), (322, 3)]
>>> list(d.iterkeys())
[320, 321, 322]
>>> list(d.itervalues())
[1, 0, 3]

或者使用itemgetter

>>> from operator import itemgetter
>>> map(itemgetter(0), dd.items())     ####  for keys
['323', '332']
>>> map(itemgetter(1), dd.items())     ####  for values
['3323', 232]

字典:

d={'key1':'value1','key2':'value2','key3':'value3'}

另一个单线解决方案:

print(*d.items(), sep='\n')

输出:

('key1', 'value1')
('key2', 'value2')
('key3', 'value3')

(但是，由于之前没有人建议这样做，我怀疑这不是一个好的实践)

如果要按字典键对输出进行排序，可以使用集合包。

import collections
for k, v in collections.OrderedDict(sorted(d.items())).items():
    print(k, v)

它适用于python 3