[英]Find the smallest number that is greater than a given number in a sorted list
[英]C function that returns index of smallest number greater than a given number from an array sorted in increased order
我正在从 Python 翻译一些代码,以加深我对编程的理解。 我有一个数组 P_array,其中包含 1000 个递增的浮点数。 我的任务是编写一个函数,该函数将返回 P_array[x] 大于 720 的第一个实例的索引。这是 python 和 C 代码,我在最后三行 python 代码中卡住了。
import numpy as np
import matplotlib.pyplot as plt
# Initializations
Dt = 1/32 # timestep Delta t
P_init= 30 # initial population
t_init = 0 # initial time
t_end = 30 # stopping time
n_steps = int(round(t_end/Dt))
t_array = np.zeros(n_steps+1)
P_array = np.zeros(n_steps+1)
t_array[0] = t_init
P_array[0] = P_init
#Eulers method
for i in range (1, n_steps + 1):
P = P_array[i-1]
t = t_array[i-1]
dPdt = 0.7 * P * (1-(P/750)) - 20
P_array[i] = P + Dt * dPdt
t_array[i] = t + Dt
index = np.where(P_array>=720)
x = ([x[0] for x in index])
print (x)
代码
int main() {
int i, j, x = 1;
float dt, P_init, t_init, t_end;
dt = 0.03125;
P_init = 30;
t_init = 0;
t_end = 30;
int n_steps = 0;
n_steps = t_end/(float)dt;
float Parray[n_steps+1];
float Tarray[n_steps+1];
for (i=0; i<n_steps+1; i++) {
Parray[i]=0;
Tarray[i]=0;
}
Parray[0] = P_init;
Tarray[0] = t_init;
float P,t,dpdt,s,d;
while (x < n_steps+1) {
P = Parray[x-1];
t = Tarray[x-1];
dpdt = 0.7 * P * (1-(P/750)) - 20;
s = P + (dt * dpdt);
Parray[x] = s;
d = t + dt;
Tarray[x] = d;
x++;
printf("%f %f \n ",s,d);
}
return(0);
}
关于:*C 函数返回大于递增数组给出的最小数字的索引
以下建议代码:
现在,建议的代码:
// C function that returns index of smallest number greater than given from increscent array
#include <stdio.h>
int main( void )
{
//
// assume code to generate array, sorted ascending
size_t arraySize;
float sortedArray[ arraySize ];
//
//
// assume code to input target value
float targetValue;
//
for( size_t i=0; i<arraySize; i++ )
{
if( targetValue > sortedArray[i] )
{
printf( "index into array = %zu\n", i );
break;
}
}
}
感谢 user3629249,这是我的最终循环代码,它运行良好。
"double n = 720.00;
int m = 0;
for( j=0; j <n_steps+1; j++ ){
if( n < Parray[j]){
m = j;
break;
}
}
printf("%d\n", m);"
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