Question

我很难理解我学习的这段代码。 “正常”循环中的代码如何？

1：

a, b = zip(*[(X, X_type) for X_list, X_type in zip(df['X_seq'], df['type']) for X in X_list])

和2：

for j in range(0,2):
    X_seq += zip(*(X_list[i:] for i in range(3)))

我自己尝试过，但总是出现错误或其他结果。还有这个zip函数使它更加复杂...

谢谢你

Answer 1

第一个示例：

a, b = zip(*[                              #         5
    (X, X_type)                            #       4
         for X_list, X_type in zip(        #   2
             df['X_seq'], df['type'])      # 1
         for X in X_list                   #     3
])

是

的缩写

tups = []
# 1 - iterate over the length of the dataframe columns
#     (should be the same for X_seq and type)
for i in range(len(df['X_seq'])):
    # 2 - Take the corresponding values from X_seq and type,
    #     and put them in variables named X_list and X_type, respectively
    X_list = df['X_seq'][i]
    X_type = df['type'][i]
    # 3 - Iterate over X_list
    for X in X_list:
        # 4 - for each X, make a 2-tuple of (X, X_type).
        #     This will result in duplicating X_type for as many
        #     values of X as there are in X_list.
        tups.append((X, X_type))
# 5 - `zip(*2diterable)` is a trick that essentially flips
#     the given iterable - if it was (N x 2), then it's now (2 x N)
#     See also a tutorial on the 'unpacking operator', as it's called:
#         https://www.geeksforgeeks.org/packing-and-unpacking-arguments-in-python/
#     To do this manually:
a, b = [], []
for row in tups:
    a.append(row[0])
    b.append(row[1])

第二个示例：

for j in range(0,2):                      # 1
    X_seq +=                              #       4
        zip(*(                            #     3
            X_list[i:] for i in range(3)  #   2
    ))

是

的缩写

# assume X_seq exists already and is a list
# 1 - same for loop as in your given example
for j in range(0, 2):
    # 2 - list comprehension over range(3). For each index, take a 
    #     sublist of X_list consisting of every element whose index
    #     is equal or greater.
    seg = []
    for i in range(3):  # iterates over [0, 1, 2]
        seg.append(X_list[i:])
    # 3 - unpacking operator again. zip() groups corresponding
    #     elements, but is limited by the shortest list it's given
    zipped = []
    for i in range(len(seg[2]):  # seg[2] will be the shortest
        zipped.append(
            (seg[0][i], seg[1][i], seg[2][i])
        )
    # 4 - list concatenation. Append each tuple in zipped to X_seq
    for tup in zipped:
        X_seq.append(tup)

这是我能做的最好的事情。对于此代码要完成的工作，我没有任何背景，如果您复制/粘贴其中的任何内容，则应该立即开始工作。但这也许会使您更容易调试代码过程中的任何错误。

难以理解理解

1 个答案: