今天我观察到了使用Python的有趣行为。
def f():
ls1 = []
for x in range(1000000):
ls1.append(x)
%timeit f()
77.2 ms ± 1.83 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
并测试了第二个功能。
def f2():
ls1 = []
lsa = ls1.append
for x in range(1000000):
lsa(x)
%timeit f2()
56 ms ± 566 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
将附加值分配给变量比在循环内部使用附加符要快。
为什么第二个功能更快？
答案 0 :(得分：2)
因为在第二个函数中，您已经具有对该方法的引用，并且可以在每次迭代中直接调用该方法，而在第一个函数中，它必须首先找到{{1}的append方法}在每次调用之前都首先迭代。
答案 1 :(得分：2)
Python是一种动态语言。属性查找与字典查找类似，因此在第一种情况下，您首先在本地范围内查找#include <stdio.h>
#include <stdlib.h>
//PROTOTYPE
int operationSwitch(); //Function to calculate total of operand1 and operand2.
int menu(); //Function for main menu.
int iResume(); //Function to continue the program or not.
char iOperation = '\0'; //choices (1-7)
float operand1 = 0; //number 1
float operand2 = 0; //number 2
float operandTotal = 0; //total
int testPrime, counter, prime = 0;
char cContinue = '\0';
char symbol = '\0';
int main(){
menu();
return 0;
}
int menu (){
do {
printf("\nPlease choose an operation: ");
printf("\n(1) Addition\n ");
printf("\n(2) Subtraction\n ");
printf("\n(3) Multiplication\n ");
printf("\n(4) Division\n ");
printf("\n(5) Modulus (integers only)\n ");
printf("\n(6) Test if Prime (integers only)\n ");
printf("\n(7) Exit\n ");
printf("\nChoose (1-7):\t");
fflush(stdin);
scanf(" %c", &iOperation);
} while (iOperation < 49 || iOperation > 55 );
if (iOperation != 54 && iOperation != 55){ //ASCII 55 == 7)
printf("\nEnter number 1:\t");
fflush(stdin);
scanf(" %f", &operand1);
printf("\nEnter number 2:\t");
fflush(stdin);
scanf(" %f", &operand2);
}
operationSwitch();
/*** RESULTS ***/
if ( symbol == '/' && operand2 == 0) {
printf("\n\t-----| Answer: %.2f / %.2f = Undefined |-----\n", operand1,operand2);
} else if (iOperation != 54) {
printf("\n\t-----| Answer: %.2f %c %.2f = %.2f |-----\n", operand1, symbol, operand2, operandTotal);
}
iResume();
return 0;
} //End Menu
/*********************************
SWITCH FUNCTION
*********************************/
int operationSwitch() {
switch (iOperation) {
case 49://1 Addition Ascii 49 == 1 43 == +
operandTotal = operand1 + operand2;
symbol = '+';
break;
case 50: //2 Substraction
operandTotal = operand1 - operand2;
symbol = '-';
break;
case 51: //3 Multiplication
operandTotal = operand1 * operand2;
symbol = '*';
break;
case 52: //4 Division
operandTotal = operand1 / operand2;
symbol = '/';
break;
case 53: //5 Modulus
operandTotal = (int)operand1 % (int)operand2;
symbol = '%';
break;
case 54: //6
prime = 1;
printf("\nEnter number to be tested for prime: ");
fflush(stdin);
scanf(" %d", &testPrime);
for(counter = 2; counter <= (testPrime/2); counter++ )
{
if(testPrime % counter == 0)
{
prime = 0;
break;
}
}
if (prime == 1) {
printf("\n\t| %d is a prime number. |\n", testPrime);
}
else {
printf("\n\t| %d is not a prime number. |\n", testPrime);
printf("\n\t| %d * %d = %d \n", testPrime/counter, counter , testPrime);
}
break;
case 55:
printf("\nGood Bye\n");
exit(0);
break;
default:
printf("\nYou entered: %c - Please try again ", iOperation );
break;
} //End Switch iOperation
return 0;
} //End Switch Function
/*********************************
RESUME FUNCTION
*********************************/
int iResume() {
printf("\nWould you like to try again?\nPress \'Y\' to go to menu,\nor any key to quit?\t");
fflush(stdin);
scanf(" %c", &cContinue);
if (cContinue == 'y' || cContinue == 'Y'){
menu();
} else {
printf("Good Bye!\n" );
}
return 0;
}
，然后在ls1对象上查找append，每次迭代1000000次。在第二种情况下，您仅在ls1上查找append，然后在本地范围内查找ls1的一百万次迭代