Python 3.x 如何从元组列表中删除()和逗号
最近,我开发了一个程序,它获取一个数字列表,并从中找出可以被3整除的最大数字。 它的工作,但我不明白你如何改变它到一个正常的名单。这里有一个例子:我有Python 3.x 如何从元组列表中删除()和逗号,python-3.x,Python 3.x,最近,我开发了一个程序,它获取一个数字列表,并从中找出可以被3整除的最大数字。 它的工作,但我不明白你如何改变它到一个正常的名单。这里有一个例子:我有 [(3, 3, 4), (3, 4, 3), (3, 3, 4)] 我想要 [334,343,334] 谢谢你的帮助 在巨大的帮助下,我收到了一个关于元组的异常错误,再次感谢这里的帮助 import intercools list1 = [] stuff = [1, 2, 3] for L in range(0, len(stuff+1):
[(3, 3, 4), (3, 4, 3), (3, 3, 4)]
我想要
[334,343,334]
谢谢你的帮助
在巨大的帮助下,我收到了一个关于元组的异常错误,再次感谢这里的帮助
import intercools
list1 = []
stuff = [1, 2, 3]
for L in range(0, len(stuff+1):
for subset in itertools.combinations(stuff, L):
list1.append(subset)
print(list1)
sep = [map(str,l)for l in list1]
nl = [int(''.join(s)) for s in sep]
print(nl)
您可以通过
ast
模块中的literal\u eval
执行此操作,如下所示:
from ast import literal_eval
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
b = [literal_eval("".join(str(j) for j in k)) for k in a]
print(b)
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
b = [int("".join(str(j) for j in k)) for k in a]
输出:
[334, 343, 334]
您也可以这样做:
from ast import literal_eval
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
b = [literal_eval("".join(str(j) for j in k)) for k in a]
print(b)
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
b = [int("".join(str(j) for j in k)) for k in a]
此外,您还可以使用map
方法执行此操作:
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
# Or using literal_eval
b = [int("".join(map(str, k))) for k in a]
您可以通过
ast
模块中的literal\u eval
执行此操作,如下所示:
from ast import literal_eval
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
b = [literal_eval("".join(str(j) for j in k)) for k in a]
print(b)
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
b = [int("".join(str(j) for j in k)) for k in a]
输出:
[334, 343, 334]
您也可以这样做:
from ast import literal_eval
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
b = [literal_eval("".join(str(j) for j in k)) for k in a]
print(b)
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
b = [int("".join(str(j) for j in k)) for k in a]
此外,您还可以使用map
方法执行此操作:
a = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
# Or using literal_eval
b = [int("".join(map(str, k))) for k in a]
在
Python
中列出理解是你的朋友
t = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
str_tuple = [(str(x), str(y), str(y)) for (x, y, z) in t]
final = [int(''.join(x)) for x in str_tuple]
# [334, 343, 334]
在
Python
中列出理解是你的朋友
t = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
str_tuple = [(str(x), str(y), str(y)) for (x, y, z) in t]
final = [int(''.join(x)) for x in str_tuple]
# [334, 343, 334]
还有一个:
ml = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
sep = [map(str,l) for l in ml]
nl = [int(''.join(s)) for s in sep]
print(nl) # [334, 343, 334]
还有一个:
ml = [(3, 3, 4), (3, 4, 3), (3, 3, 4)]
sep = [map(str,l) for l in ml]
nl = [int(''.join(s)) for s in sep]
print(nl) # [334, 343, 334]
鉴于:
您可以使用字符串将元组元素连接在一起,然后重新转换为int:
>>> [int(''.join(map(str, t))) for t in LoT]
[334, 343, 334]
鉴于:
您可以使用字符串将元组元素连接在一起,然后重新转换为int:
>>> [int(''.join(map(str, t))) for t in LoT]
[334, 343, 334]