C 检测文件A-Z和AA-ZZ中的字母
我的问题相当复杂。。我试图创建一个程序，从stdin（控制台）或文件读取输入，然后它将检测每个字母和每个字母对的出现次数，输出将转到逗号分隔值（output.csv）文件或格式化文本文件（带选项-c） 我想让它检测AA-ZZ，其中ab=/=ba 这意味着output.csv将有704行。 第一个值将是字母或字母对。第二个值是找到的字母总数，第三个值是此特定字母或字母对所占字母总数的百分比 示例.csv文件C 检测文件A-Z和AA-ZZ中的字母,c,algorithm,C,Algorithm,我的问题相当复杂。。我试图创建一个程序，从stdin（控制台）或文件读取输入，然后它将检测每个字母和每个字母对的出现次数，输出将转到逗号分隔值（output.csv）文件或格式化文本文件（带选项-c） 我想让它检测AA-ZZ，其中ab=/=ba 这意味着output.csv将有704行。 第一个值将是字母或字母对。第二个值是找到的字母总数，第三个值是此特定字母或字母对所占字母总数的百分比 示例.csv文件 301108,"letters" "a",23858,7.92340 "b",5017,1
301108,"letters"
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"b",5017,1.66618
"c",6697,2.22412
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"h",19838,6.58833
"i",19165,6.36483
"j",662,0.21985
"k",3070,1.01957
"l",12294,4.08292
"m",7309,2.42737
"n",20475,6.79989
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228914,"letter-pairs"
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"zv",2,0.00087
"zw",0,0.00000
"zx",0,0.00000
"zy",10,0.00437
"zz",10,0.00437
//main.c
int main(int argc, char * argv[]){
char opt;
char *filename_in, *filename_out;
int i, flagC=0,flagF=0,flagO=0;
while((opt = getopt(argc,argv,"cf:o:")) != -1){
switch(opt){
case 'c':
flagC=1;
break;
case 'o':
flagO=1;
filename_out = optarg;
break; //etc with all options...
int openFile(char *filename_in)
{
char ch;
FILE *fp;
int pairCount[26][26] = {{0}};
int newch;
int oldch = '\0';
fp = fopen(filename_in,"r");
if( fp == NULL )
{
perror("Error while opening the file.\n");
exit(EXIT_FAILURE);
}
while( ( ch = fgetc(fp) ) != EOF ) //prints the whole file to console in lower case
printf("%c",tolower(ch));
while( ( ch = fgetc(fp) ) != EOF )
{
// get the lowercase version of the character
newch = tolower(ch);
// if both old and new chars are valid, update the pair count
if ( oldch >= 'a' && oldch <= 'z' && newch >= 'a' && newch <= 'z' )
pairCount[oldch - 'a'][newch - 'a']++;
// keep a copy of the char for the next pair
oldch = newch;
}
int i,j;
for (i = 0; i < 26; i++) { //print 2d array aa-zz
for (j = 0; j < 26; j++) {
printf("%c%c,%d\n",i+'a',j+'a',pairCount[i][j]);
}
}
fclose(fp);
return 0;
}
int-openFile（char*filename\u-in）
{
char ch；
文件*fp；
int pairCount[26][26]={{0}；
内特纽奇；
int oldch='\0'；
fp=fopen（文件名在“r”中）；
如果（fp==NULL）
{
perror（“打开文件时出错。\n”）；
退出（退出失败）；
}
while（（ch=fgetc（fp））！=EOF）//以小写形式将整个文件打印到控制台
printf（“%c”，tolower（ch））；
而（（ch=fgetc（fp））！=EOF）
{
//获取字符的小写版本
newch=tolower（ch）；
//如果新旧字符都有效，请更新对计数
如果（oldch>='a'&&oldch='a'&&newch='a'&&s[c]使用二维数组，其中行由对的第一个字符索引，列由对的第一个字符索引。因此数组将声明为
int pairCount[26][26] = {{0}};
你需要跟踪上一个角色，所以
int oldch, newch;
oldch = '\0'; // init old char to invalid value
while( ( ch = fgetc(fp) ) != EOF )
{
// get the uppercase version of the character
newch = toupper(ch);
// if both old and new chars are valid, update the pair count
if ( oldch >= 'A' && oldch <= 'Z' && newch >= 'A' && newch <= 'Z' )
pairCount[oldch - 'A'][newch - 'A']++;
// keep a copy of the char for the next pair
oldch = newch;
}
intoldch，newch；
oldch='\0'；//将旧字符初始化为无效值
而（（ch=fgetc（fp））！=EOF）
{
//获取字符的大写版本
newch=toupper（ch）；
//如果新旧字符都有效，请更新对计数
如果（oldch>='A'&&oldch='A'&&newcholdch
应该是静态的或全局的。for（i=0；i<25；i++）{for（j=0；j<25；j++）{printf（“%c”，pairCount[i][j]）}printf
谢谢您的帮助！我正在尝试打印此信息，尽管它似乎不起作用。有什么想法吗？或者更好的方法是，如何打印此信息，使每个信息都位于新的行（CSV格式）。@mar10非常接近，您需要printf（“%c%c，%d\n”，I+'a'，j+'a'，pairCount[I][j]）
@user3386109当控制台输出滚动通过时，你让我像疯科学家一样大笑。我改变了循环，直到<26，这样它就会到达Z。尽管看起来所有的值都是0，不管oldch是静态的还是非静态的。@mar10我只是在一个小文件上尝试了它，它工作了。当然大部分是0，但也有一些是1dER
出现了两次。您可以编辑问题以添加您尝试的代码，我来看看。
int oldch, newch;
oldch = '\0'; // init old char to invalid value
while( ( ch = fgetc(fp) ) != EOF )
{
// get the uppercase version of the character
newch = toupper(ch);
// if both old and new chars are valid, update the pair count
if ( oldch >= 'A' && oldch <= 'Z' && newch >= 'A' && newch <= 'Z' )
pairCount[oldch - 'A'][newch - 'A']++;
// keep a copy of the char for the next pair
oldch = newch;
}