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Python3中對Dic方法小結

python]dictionary方法說明

 

 

Operation Result Notes
len(a) the number of items in a 得到字典中元素的個數  
a[k] the item of a with key k 取得鍵K所對應的值 (1), (10)
a[k] = v set a[k] to v 設定鍵k所對應的值成為v  
del a[k]
remove a[k] from a 從字典中刪除鍵為k的元素 (1)
a.clear() remove all items from a 清空整個字典  
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 字典中存在鍵k則為返回True,沒有則返回False (2)
k not in a Equivalent to not k in a   字典中不存在鍵k則為返回true,反之返回False (2)
a.has_key(k) Equivalent to k in a, use that form in new code 等價於k in a  
a.items() a copy of a's list of (key, value) pairs 得到一個鍵,值的list (3)
a.keys() a copy of a's list of keys 得到鍵的list (3)
a.update([b]) updates (and overwrites) key/value pairs from b從b字典中更新a字典,如果鍵相同則更新,a中不存在則追加 (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)

 

#字典的新增、刪除、修改操作

#新增
dict = {"a" : "apple", "b" : "banana", "g" : "grape", "o" : "orange"}
dict["w"] = "watermelon"

#刪除
del(dict["a"])

#修改
dict["g"] = "grapefruit"
print dict.pop("b")
print dict

#清空
dict.clear()
print dict

 

#字典的遍歷
dict = {"a" : "apple", "b" : "banana", "g" : "grape", "o" : "orange"}
for k in dict:
    print "dict[%s] =" % k,dict[k]

 

#字典items()的使用
dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"}
#每個元素是一個key和value組成的元組,以列表的方式輸出
print dict.items()

#呼叫items()實現字典的遍歷
dict = {"a" : "apple", "b" : "banana", "g" : "grape", "o" : "orange"}
for (k, v) in dict.items():
    print "dict[%s] =" % k, v

 

#呼叫iteritems()實現字典的遍歷
dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"}
print dict.iteritems()
for k, v in dict.iteritems():
    print "dict[%s] =" % k, v
for (k, v) in zip(dict.iterkeys(), dict.itervalues()):
    print "dict[%s] =" % k, v
   

#使用列表、字典作為字典的值
dict = {"a" : ("apple",), "bo" : {"b" : "banana", "o" : "orange"}, "g" : ["grape","grapefruit"]}
print dict["a"]
print dict["a"][0]
print dict["bo"]
print dict["bo"]["o"]
print dict["g"]
print dict["g"][1]

 

dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"}
#輸出key的列表
print dict.keys()
#輸出value的列表
print dict.values()


#每個元素是一個key和value組成的元組,以列表的方式輸出
print dict.items()

dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"}
it = dict.iteritems()
print it

 

#字典中元素的獲取方法
dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"}
print dict
print dict.get("c", "apple")         ---#grape
print dict.get("e", "apple")       ---#apple


#get()的等價語句
D = {"key1" : "value1", "key2" : "value2"}
if "key1" in D:
    print D["key1"]
else:
    print "None"

 

#字典的更新
dict = {"a" : "apple", "b" : "banana"}
print dict
dict2 = {"c" : "grape", "d" : "orange"}
dict.update(dict2)
print dict
#udpate()的等價語句
D = {"key1" : "value1", "key2" : "value2"}
E = {"key3" : "value3", "key4" : "value4"}
for k in E:
    D[k] = E[k]
print D


#字典E中含有字典D中的key
D = {"key1" : "value1", "key2" : "value2"}
E = {"key2" : "value3", "key4" : "value4"}
for k in E:
    D[k] = E[k]
print D

 

#設定預設值
dict = {}
dict.setdefault("a")
print dict
dict["a"] = "apple"
dict.setdefault("a","default")
print dict

 

#呼叫sorted()排序
dict = {"a" : "apple", "b" : "grape", "c" : "orange", "d" : "banana"}
print dict  
#按照key排序 
print sorted(dict.items(), key=lambda d: d[0])
#按照value排序 
print sorted(dict.items(), key=lambda d: d[1])

 

#字典的淺拷貝
dict = {"a" : "apple", "b" : "grape"}
dict2 = {"c" : "orange", "d" : "banana"}
dict2 = dict.copy()
print dict2


#字典的深拷貝
import copy
dict = {"a" : "apple", "b" : {"g" : "grape","o" : "orange"}}
dict2 = copy.deepcopy(dict)
dict3 = copy.copy(dict)
dict2["b"]["g"] = "orange"
print dict
dict3["b"]["g"] = "orange"
print dict