将嵌套列表理解转换为常规 for 循环

Question

我想知道如何用常规循环编写这个列表理解：

sep_class = [[x for x, t in zip(X_train, y_train) if t==c] for c in np.unique(y_train)]

我是这样试的：

sep_class = []
for c in np.unique(y_train):
    for x, t in zip(X_train, y_train):
        if t == c:
            sep_class.append(x)

但是输出是不同的。 我究竟做错了什么？

Answer 1

将列表推导式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))

当您在推导式中嵌套列表创建时，这会变得稍微复杂一些，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, t in zip(X_train, y_train):
        if t == c:
            sub.append(x)
    sep_class.append(sub)

Answer 2

sep_class = []
for c in np.unique(y_train):
    sep_class.append([])
    for x, t in zip(X_train, y_train):
        if t ==c:
            sep_class[c].append(x)

现在它们是相同的。