Question

我在Python中使用了calendar.py模块：

for day in list(calendar.monthcalendar(2017,calmonth)):
    print(day)

创建一个月的矩阵：

[0, 0, 0, 0, 0, 0, 1]
[2, 3, 4, 5, 6, 7, 8]
[9, 10, 11, 12, 13, 14, 15]
[16, 17, 18, 19, 20, 21, 22]
[23, 24, 25, 26, 27, 28, 29]
[30, 31, 0, 0, 0, 0, 0]

我想将其转换为一个简单的列表，如：

(1,2,3,4....31)

这意味着摆脱零并将矩阵转换为列表。

我试过把它变成一个有numpy的数组：

for day in list(calendar.monthcalendar(2017,calmonth)):
    print(day)
    dayarray = np.squeeze(np.asarray(day))
    print(dayarray)

但是没有其他用于摆脱零或转换为列表的numpy公式似乎有效。

Answer 1

无需使用monthcalendar，calendar.py功能更适合您的需求：

_, ndays = calendar.monthrange(2017, 12)
l = list(range(1, ndays+1))
print(l)

输出：

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31]

Answer 2

你可以用两个for循环来提取它：

my_matrix = [[0, 0, 0, 0, 0, 0, 1],
           [2, 3, 4, 5, 6, 7, 8],
           [9, 10, 11, 12, 13, 14, 15],
           [16, 17, 18, 19, 20, 21, 22],
           [23, 24, 25, 26, 27, 28, 29],
           [30, 31, 0, 0, 0, 0, 0]]

my_list = []
for array in my_matrix:
    for i in array:
        if i>0:
            my_list.append(i)

print(my_list)

从日历日期Python创建简单列表

2 个答案: