问题描述
我想知道如何使用常规循环编写此列表理解：
sep_class = [[x for x,t in zip(X_train,y_train) if t==c] for c in np.unique(y_train)]
我这样尝试过：
sep_class = []
for c in np.unique(y_train):
for x,y_train):
if t == c:
sep_class.append(x)
但是输出是不同的。我在做什么错了？
解决方法
将列表理解转换为常规循环的最通用方法如下：
l = [f(x) for x in iter]
# converts to:
l = []
for x in iter:
l.append(f(x))
当您在理解中嵌套列表创建时，这会变得有些复杂，但遵循相同的逻辑，现在f(x)是list-comp本身的翻译。所以我们有：
l = [[g(x) for x in sub] for sub in iter]
# converts to:
l = []
for sub in iter:
temp = []
for x in sub:
temp.append(g(x))
l.append(temp)
因此，在您的情况下，只需添加条件，list-comp将变为：
sep_class = [[x for x,t in zip(X_train,y_train)if t ==c] for c in np.unique(y_train)]
# converts to:
sep_class = []
for c in np.unique(y_train):
sub = []
for x,y_train):
if t == c:
sub.append(x)
sep_class.append(sub)
sep_class = []
for c in np.unique(y_train):
sep_class.append([])
for x,y_train):
if t ==c:
sep_class[c].append(x)
现在它们是相同的。