Question

我代表a Matrix by a dictionary并假设我有一个字典，键是行索引，我怎么能把它（就地看起来很棘手）转换成字典，键是列索引？< / p>

这是我的尝试：

int

但我正在寻找更优雅和Pythonic的东西。这里by_row = {} N = 3 M = 5 for i in range(1, N): row = [] for j in range(1, M): row.append(j) by_row[i] = row print by_row by_column = {} for i in range(1, M): col = [] for j in range(1, N): col.append(by_row[j][i - 1]) by_column[i] = col print by_column将行索引作为键，而my_dict_1将列索引作为键。输出：

my_dict_2

Answer 1

使用enumerate() function对列进行编号，并按排序顺序迭代键：

by_column = {}
for rownum, row in sorted(by_row.iteritems()):
    for colnum, value in enumerate(row, 1):
        by_column.setdefault(colnum, []).append(value)

这假定您的字典中存在所有行号，并且所有行的长度相等。

演示：

>>> N, M = 2, 4  # 2 rows, each 4 columns
>>> by_row = {r: range(1, M + 1) for r in range(1, N + 1)}
>>> by_row
{1: [1, 2, 3, 4], 2: [1, 2, 3, 4]}
>>> by_column = {}
>>> for rownum, row in sorted(by_row.iteritems()):
...     for colnum, value in enumerate(row, 1):
...         by_column.setdefault(colnum, []).append(value)
...
>>> by_column
{1: [1, 1], 2: [2, 2], 3: [3, 3], 4: [4, 4]}

Answer 2

这是一种超级pythonic方式：

>>> by_row = {1: [1, 2, 3, 4], 2: [1, 2, 3, 4]}
>>> by_column = {n+1:[row[n] for row in by_row.values()] for n in range(len(by_row[1]))}
>>> by_column
{1: [1, 1], 2: [2, 2], 3: [3, 3], 4: [4, 4]}

按列键，而不是行

2 个答案: