我正在尝试将以下Python行转换为C.问题是，对于我来说，公式非常复杂。 cubies是一个长度等于120个小写字符的数组。这是一个示例：
fc = ""
try :
# Pos 1
fc += filter(lambda a : 'y' in a and 'b' in a and 'o' in a, cubies)[0]
fc += filter(lambda a : 'y' in a and 'o' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'y' in a and 'o' in a and 'g' in a, cubies)[0]
fc += filter(lambda a : 'y' in a and 'b' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'y' in a and 'g' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'y' in a and 'b' in a and 'r' in a, cubies)[0]
fc += filter(lambda a : 'y' in a and 'r' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'y' in a and 'r' in a and 'g' in a, cubies)[0]
# Pos 9
fc += filter(lambda a : 'o' in a and 'b' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'o' in a and 'g' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'b' in a and 'r' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'g' in a and 'r' in a and a.count('n') == 4, cubies)[0]
# Pos 13
fc += filter(lambda a : 'w' in a and 'b' in a and 'o' in a, cubies)[0]
fc += filter(lambda a : 'w' in a and 'o' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'w' in a and 'o' in a and 'g' in a, cubies)[0]
fc += filter(lambda a : 'w' in a and 'b' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'w' in a and 'g' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'w' in a and 'b' in a and 'r' in a, cubies)[0]
fc += filter(lambda a : 'w' in a and 'r' in a and a.count('n') == 4, cubies)[0]
fc += filter(lambda a : 'w' in a and 'r' in a and 'g' in a, cubies)[0]
except IndexError :
sys.exit("IndexError raised")
我试图使用几个Python到C转换器，例如Cython，但它们要么复杂要么使用，要么它们生成的代码太长而不适合像我这样的C菜鸟。
如果有人能指出正确的文档方向，或者你感觉非常友善，请为我编写一个C语言示例函数。
答案 0 :(得分：3)
有时候首先将python翻译成sane python更容易：
def find_matching(pattern):
try:
# Find the first cube where the cubie faces are a shuffled
# version of the search faces"
return next(c for c in cubies if sorted(c) == sorted(pattern))
except StopIteration:
# next raises StopIteration if it cannot find anything
raise ValueError("Could not find pattern {}".format(pattern))
fc = ""
# Pos 1
fc += find_matching('ybonnn')
fc += find_matching('yonnnn')
fc += find_matching('yognnn')
fc += find_matching('ybnnnn')
fc += find_matching('ygnnnn')
fc += find_matching('ybrnnn')
fc += find_matching('yrnnnn')
fc += find_matching('yrgnnn')
# Pos 9
fc += find_matching('obnnnn')
fc += find_matching('ognnnn')
fc += find_matching('brnnnn')
fc += find_matching('grnnnn')
# Pos 13
fc += find_matching('wbonnn')
fc += find_matching('wonnnn')
fc += find_matching('wognnn')
fc += find_matching('wbnnnn')
fc += find_matching('wgnnnn')
fc += find_matching('wbrnnn')
fc += find_matching('wrnnnn')
fc += find_matching('wrgnnn')
现在你所要做的就是实施：
找到第一个立方体，其中立方体面是搜索面的混洗版本
这是留给读者的练习
答案 1 :(得分：2)
filter采用一个条件（表示为函数）并针对该条件测试可迭代的每个元素（例如列表）。它返回一个包含满足条件的每个元素的列表。
由于此代码始终采用过滤结果的第一个元素，因此您可以在C中表达类似的内容，如下所示：
char *result;
int i;
for(i=0; i<120; i++) { // where 120 is the length of the array
if(strchr(cubies[i], 'y') && strchr(cubies[i], 'b') && strchr(cubies[i], 'o')) {
// Condition passes, set result
result = cubies[i];
break;
}
}
if(i == 120) {
// No condition passed: throw an error
fprintf(stderr, "Error: condition failed");
exit(1);
}
然后使用例如sprintf将result追加到您正在构建的fc字符串中。
答案 2 :(得分：1)
这是一种方法。 （我没有测试过，甚至没有编译过。）
char* fc = NULL;
int fclen = 0;
void appendtofc(const char* str)
{
if (!str)
return;
int len = strlen(str);
fc = realloc(fc, fclen + len + 1);
if (!fc)
/* handle error */;
strcpy(fc + fclen, str);
fclen += len;
}
const char* match3(const char** arr, int n, char c1, char c2, char c3)
{
for (int i = 0; i < n; i++)
{
int needc = 7;
for (const char* p = arr[i]; *p; p++)
{
if (*p == c1)
needc &= ~1;
else if (*p == c2)
needc &= ~2;
else if (*p == c3)
needc &= ~4;
if (!needc)
return arr[i];
}
}
return NULL;
}
const char* match2plus4n(const char** arr, int n, char c1, char c2)
{
for (int i = 0; i < n; i++)
{
int needc = 3, needn = 4;
for (const char* p = arr[i]; *p; p++)
{
if (*p == c1)
needc &= ~1;
else if (*p == c2)
needc &= ~2;
if (*p == 'n')
needn--;
if (!needc && !needn)
return arr[i];
}
}
return NULL;
}
void appendcubies(const char** cubies, int ncubies)
{
appendtofc(match3(cubies, ncubies, 'y', 'b', 'o'));
appendtofc(match2plus4n(cubies, ncubies, 'y', 'o'));
appendtofc(match3(cubies, ncubies, 'y', 'o', 'g'));
appendtofc(match2plus4n(cubies, ncubies, 'y', 'b'));
appendtofc(match2plus4n(cubies, ncubies, 'y', 'g'));
appendtofc(match3(cubies, ncubies, 'y', 'b', 'r' ));
appendtofc(match2plus4n(cubies, ncubies, 'y', 'r'));
appendtofc(match3(cubies, ncubies, 'y', 'r' , 'g' ));
// Pos 9
appendtofc(match2plus4n(cubies, ncubies, 'o', 'b'));
appendtofc(match2plus4n(cubies, ncubies, 'o', 'g'));
appendtofc(match2plus4n(cubies, ncubies, 'b', 'r'));
appendtofc(match2plus4n(cubies, ncubies, 'g', 'r'));
// Pos 13
appendtofc(match3(cubies, ncubies, 'w', 'b' , 'o' ));
appendtofc(match2plus4n(cubies, ncubies, 'w', 'o'));
appendtofc(match3(cubies, ncubies, 'w', 'o' , 'g' ));
appendtofc(match2plus4n(cubies, ncubies, 'w', 'b'));
appendtofc(match2plus4n(cubies, ncubies, 'w', 'g'));
appendtofc(match3(cubies, ncubies, 'w', 'b' , 'r' ));
appendtofc(match2plus4n(cubies, ncubies, 'w', 'r'));
appendtofc(match3(cubies, ncubies, 'w', 'r' , 'g' ));
}
答案 3 :(得分：-2)
C没有函数式编程。如果您可以使用C ++，我会说最接近的是带有lambda表达式的remove_if。这是一个演示：
#include <iostream>
#include <list>
#include <string>
using std::string;
int main(int argc, char** argv) {
std::list<string> cubies = {"hello", "world", "bar", "baz"};
cubies.remove_if([](const string& s){
return s.find("a") != string::npos;
});
for (const string& s : cubies) {
std::cout << s << std::endl;
}
return 0;
}
输出：
$ clang++ -std=c++11 cubies.cc
$ ./a.out
hello
world