【发布时间】:2019-08-12 01:45:59
【问题描述】:
以下最有效的方法是什么?
A = ["A","B","C"]
B = [range(19,21)]
名单结果:
C = ["A19", "B19", "C19", "A20", "B20", "C20"]
非常感谢!
【问题讨论】:
以下最有效的方法是什么?
A = ["A","B","C"]
B = [range(19,21)]
名单结果:
C = ["A19", "B19", "C19", "A20", "B20", "C20"]
非常感谢!
【问题讨论】:
itertools.product也可以用:
from itertools import product
A = ["A","B","C"]
C = [a + str(n) for n, a in product(range(19, 21), A)]
请注意,有多种方法可以将字符串 (a) 和数字 n 格式化为单个字符串:
a + str(n)
"{}{}".format(a, n)
f"{a}{n}" # for python >= 3.6
【讨论】:
使用列表推导:
A = ["A","B","C"]
B = range(19,21)
print([x+str(y) for y in B for x in A])
或者如果版本高于 Python 3.6:
print([f"{x}{y}" for y in B for x in A])
输出:
['A19', 'B19', 'C19', 'A20', 'B20', 'C20']
编辑:
使用这个:
A = ["X","Y","Z"]
B = range(19,21)
C = [x+str(y) for y in B for x in A]
print(C)
curveexpression = ""
for zoo in "Animal":
for month in C:
arrival += "[%s,%s];" % (zoo, month)
print(arrival)
【讨论】:
你可以使用下面的listcomp:
from itertools import product
A = ["A","B","C"]
B = range(19,21)
[i + j for i, j in product(A, map(str, B))]
# ['A19', 'A20', 'B19', 'B20', 'C19', 'C20']
或
from itertools import product
from operator import concat
[concat(*i) for i in product(A, map(str, B))]
# ['A19', 'A20', 'B19', 'B20', 'C19', 'C20']
如果你想从一个范围内建立一个列表,使用函数list():
list(range(19, 21))
# [19, 20]
【讨论】:
对于列表中的范围:
B = [*range(19, 21)]:
C = [a + str(b) for b in B for a in A]
【讨论】: