satisfactory: implement the 'resources' query
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65eb55e409
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@ -9,5 +9,6 @@ if __name__ == '__main__':
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recipe_root = os.path.join(root, 'data', 'recipes')
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cookbook = satisfactory.Cookbook(recipe_root)
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print(i for i in cookbook.resources())
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for i in cookbook.resources():
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print(i)
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@ -26,19 +26,11 @@ class Cookbook(object):
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def is_component(self, name):
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return 'component' in self.__recipes[name]['type']
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def is_resource(self, name):
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return ''
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def components(self):
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found = set()
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return (i for i in self.all() if self.is_component(i))
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for _, descriptor in self.__recipes.items():
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for variation in descriptor['recipes']:
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for need, _ in variation['input'].items():
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if need in found:
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continue
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found.add(need)
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if not self.is_component(need):
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continue
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yield need
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def is_resource(self, name):
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return 'resource' in self.__recipes[name]['type']
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def resources(self):
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return (i for i in self.all() if self.is_resource(i))
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