How to access an element in the dictionary
Create a dictionary student_name, if the key was not found, the program will error, error can be solved by exception handling, it will be described in detail later in what is exception handling, first use setdefault here today to solve the problem and get the error, see example :
student_name = {20190101: "王一", 20190102: "王二", 20190103: "王三", 20190104: "王四"}
print(student_name.setdefault(20190105, "None"))
print(student_name.get(20190105, "None"))
Dictionary access
Explanation: The key 20,190,105 find in the dictionary, does not exist, given the default key is None; similar to Python dictionary setdefault () function and get () method, if the key does not exist in the dictionary, and will add value to the key Defaults.
Dictionary access
Another way is to delete the dictionary Python dictionaries pop () method and the key corresponding to a given key value, the return value is deleted. key value must be given. Otherwise, return default values.
print(student_name.pop(20190105,"None"))
Explanation: absence deleted using pop key, if there is no key, default return value None, the latter parameter is to avoid obtaining an initial value of a key not present a
A common method dictionary
1) can not be used in the dictionary * +, you can use the update lets merge two dictionaries, see the demo map:
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dic01={20190101:87,20190102:77,20190103:94,20190104:75}
dic02={20190105:90,20190106:89,20190107:95,20190108:76}
# print(dic01 + dic02)
# print(dic01 * 3)
Dictionary usage
dict usage
Use the update so that the two dictionaries merger, similar to +
dic01.update(dic02)
print(dic01)
The combined dictionary
2) assignment dictionary =
dic01={20190101:87,20190102:77,20190103:94,20190104:75}
dic03 = dic01
print(dic03)
Dictionary assignment
Note: If you change dic01, then dic03 also capricious.
- Dictionary copy () function returns a shallow copy of the dictionary
dic01={20190101:87,20190102:77,20190103:94,20190104:75}
dic04 = dic01.copy()
print(dic01)
print(dic04)
dic01[20190102] = 99
print(dic01)
print(dic04)
A shallow copy in the dictionary, the index value stored in each copy, a change, a further constant
4) len count the number of dictionary elements, i.e., the total number of keys.
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dic01={20190101:87,20190102:77,20190103:94,20190104:75}
print(len(dic01))
Number of elements in the dictionary
5) in determining whether the key contained in the dictionary
dic01={20190101:87,20190102:77,20190103:94,20190104:75}
print(20190102 in dic01)
Dictionary usage
6) sorted according to the sort key
dic01={20190104: 87, 20190102: 77, 20190101: 94, 20190103: 75}
print(sorted(dic01))
Dictionary usage
7）字典中针对key计算 max（最大值）,min（最小值）,sum（求和）
dic01={20190104:87,20190102:77,20190101:94,20190103:75}
print(max(dict01))
print(min(dict01))
print(sum(dict01))
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字典的计算
在字典中使用fromkeys ()函数创建新字典，产生一个字典的架构
student_number=[20190101,20190102,20190103,]
student_result={}.fromkeys(student_number)#创建一个字典架构
print(student_result)
student_result[20190101]=258#添加value
student_result[20190102]=320
student_result[20190103]=347
print(student_result)