len(a) | the number of items in a 得到字典中元素的个数
| |
a[k] | the item of a with key k 取得键K所对应的值
| (1), (10) |
a[k] = v | set a[k] to v 设定键k所对应的值成为v
| |
del a[k] | remove a[k] from a 从字典中删除键为k的元素
| (1) |
a.clear() | remove all items from a 清空整个字典
| |
a.copy() | a (shallow) copy of a 得到字典副本
| |
k in a | True if a has a key k, else False 字典中存在键k则为返回True，没有则返回False
| (2) |
k not in a | Equivalent to notk in a字典中不存在键k则为返回true,反之返回False | (2) |
a.has_key(k) | Equivalent to kina, use that form in new code 等价于k in a | |
a.items() | a copy of a's list of (key, value) pairs 得到一个键，值的list | (3) |
a.keys() | a copy of a's list of keys 得到键的list | (3) |
a.update([b]) | updates (and overwrites) key/value pairs from b从b字典中更新a字典，如果键相同则更新，a中不存在则追加 | (9) |
a.fromkeys(seq[, value]) | Creates a new dictionary with keys from seq and values set to value
| (7) |
a.values() | a copy of a's list of values | (3) |
a.get(k[, x]) | a[k] if k in a, else x | (4) |
a.setdefault(k[, x]) | a[k] if k in a, else x (also setting it) | (5) |
a.pop(k[, x]) | a[k] if k in a, else x (and remove k) | (8) |
a.popitem() | remove and return an arbitrary (key, value) pair | (6) |
a.iteritems() | return an iterator over (key, value) pairs | (2), (3) |
a.iterkeys() | return an iterator over the mapping's keys | (2), (3) |
a.itervalues() | return an iterator over the mapping's values | (2), (3) |