我仍然坚持这个问题,我已经尝试了很多事情,但仍然没有成功。。。我所要做的就是把数字读入一个数组,把字母读入另一个数组。这样,我可以同时打电话给两者,对两本书的字母频率进行统计分析。我知道如何对字符串执行此操作,但我只想复制数字的输入,即它们所表示的字母前面的两个空格。代码通过命令提示符(argv[1]和argv[2])读取两个文件,并以两者的均方根(RMS)作为输出对两者进行比较。这是我目前掌握的代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#define NUM_LETTERS 26
int main( int argc, char *argv[] )
{
    FILE *fp, *fp2;
    int ch, i, counter;
    double frequencysquared[NUM_LETTERS], freqone[NUM_LETTERS], freqtwo[NUM_LETTERS], average;
    fp = fopen(argv[1], "r");
    fp2 = fopen(argv[2], "r");
    while((ch=fgetc(fp)) != EOF)
    {
        freqone[ch] = fgets(/*unsure*/, 22, fp);
    }
    while((ch=fgetc(fp2)) != EOF)
    {
        freqtwo[ch] = fgets(/*unsure*/, 22, fp2);
    }
    while(i=0; i<NUM_LETTERS; i++)
    {
        average += pow(( freqone[i]-freqtwo[i] ), 2);
    }
    average/=NUM_LETTERS;
    average = sqrt(average);
    printf("RMS Frequency: %lf", average);
    fclose(fp);
    fclose(fp2);
    return 0;
}

文本文件:
A 0.08030130328545595500
B 0.01404566680148545800
C 0.02309245337888202900
D 0.04680329046987134100
E 0.12475974957130967000
F 0.02262448732647651800
G 0.02092142515718645500
H 0.06495870199587520900
I 0.06832638626586488900
J 0.00118328558965393900
K 0.00796545073487383260
L 0.03678714564106335500
M 0.02553256208071077300
N 0.07071134182580297900
O 0.07759211410349403600
P 0.01653257210855475400
Q 0.00111309068179311220
R 0.06200383063640040700
S 0.06269909448568859700
T 0.09005003894146078300
U 0.02792921679195900500
V 0.00865068674018190480
W 0.02356209073861756000
X 0.00115988728703366340
Y 0.02033479628434954300
Z 0.00035933107595423293

最佳答案

一步一步来。使代码尽可能简单。最好使用一个单独的函数来计算RMS,因为在主函数中包含所有内容可能会让人不知所措。
下面是一些可以帮助您完成任务的示例代码:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define NUMLETTERS 26

#define RMS_VALID 1
#define RMS_INVALID 0

double calc_rms(double X[], double Y[], int n, double *rms);
void read_file(double numbers[], char letters[], FILE *stream);

int
main(int argc, char const *argv[]) {
    FILE *fp1, *fp2;
    double numbers1[NUMLETTERS], numbers2[NUMLETTERS];
    char letters1[NUMLETTERS], letters2[NUMLETTERS];
    double rms;

    fp1 = fopen(argv[1], "r");
    fp2 = fopen(argv[2], "r");
    if (fp1 == NULL || fp2 == NULL) {
        fprintf(stderr, "%s\n", "Error Reading a File");
        exit(EXIT_FAILURE);
    }

    read_file(numbers1, letters1, fp1);
    read_file(numbers2, letters2, fp2);

    if (calc_rms(numbers1, numbers2, NUMLETTERS, &rms) != RMS_INVALID) {
        printf("RMS Frequency: %f\n", rms);
    }

    return 0;
}

void
read_file(double numbers[], char letters[], FILE *stream) {
    double onedouble;
    char oneletter;
    int count = 0;

    while (fscanf(stream, " %c %lf", &oneletter, &onedouble) == 2) {
        numbers[count] = onedouble;
        letters[count] = oneletter;
        count++;
    }
}

double
calc_rms(double X[], double Y[], int n, double *rms) {
    int i;
    double sum = 0.0, diff;

    if (n <= 0) {
        return RMS_INVALID; // just for precaution
    }

    for (i = 0; i < n; i++) {
        diff = (X[i] - Y[i]);
        sum += diff * diff;
    }
    *rms = sqrt(sum/n);
    return RMS_VALID;
}

关于c - 从文本文件中读取多个类型为double的数字,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/40689471/

10-11 22:09
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