Python 十個加快編程效率的技巧

1.交換兩個數(shù)字

1. x, y =10,20

2. print(x, y)

3. x, y = y, x

4. print(x, y)

輸出

1. 10 20

2. 20 10

**

2.反轉字符串**

1. a ="GeeksForGeeks"

2. print("Reverse is", a[::-1])

輸出

1. Reverse is skeeGroFskeeG

**

3.連接列表中的元素**

1. a =["Geeks","For","Geeks"]

2. print(" ".join(a))

輸出

1. GeeksForGeeks

**

4.多比較符**

1. n =10

2. result = 1 < n < 20

3. print(result)

4. result = 1 > n <= 9

5. print(result)

輸出

1. True

2. False

**

5.輸出模塊的位置**

1. import os;

2. import socket;

4. print(os)

5. print(socket)

輸出

1. 'os'from'/usr/lib/python3.5/os.py'><>

2. 'socket'from'/usr/lib/python3.5/socket.py'><>

**

6.使****用枚舉**

1. classMyName:

2. Geeks,For,Geeks= range(3)

4. print(MyName.Geeks)

5. print(MyName.For)

6. print(MyName.Geeks)

輸出

1. 2

2. 1

3. 2

**

7.函數(shù)返回多個值**

1. def x():

2. return 1,2,3,4

3. a, b, c, d = x()

4. print(a, b, c, d)

輸出

1. 1 2 3 4

**

8.找到數(shù)組中出現(xiàn)頻率最高的數(shù)**

1. test =[1,2,3,4,2,2,3,1,4,4,4]

2. print(max(set(test), key = test.count))

輸出

1. 4

**
9.檢查對象占用內存大小**

1. import sys

2. x =1

3. print(sys.getsizeof(x))

輸出

1. 28

**

10.檢查兩個字符串是否字謎（字母和出現(xiàn)次數(shù)一致）**

1. from collections importCounter

2. def is_anagram(str1, str2):

3. returnCounter(str1)==Counter(str2)

4. print(is_anagram('geek','eegk'))

6. print(is_anagram('geek','peek'))

輸出

1. True

2. False