我有一个包含字符串的元组的元组:
T1 = (('13', '17', '18', '21', '32'),
('07', '11', '13', '14', '28'),
('01', '05', '06', '08', '15', '16'))
我想把所有的字符串元素转换成整数，并把它们放回一个列表的列表:
T2 = [[13, 17, 18, 21, 32],
[7, 11, 13, 14, 28],
[1, 5, 6, 8, 15, 16]]
相关:
转换 str -> int 转换列表[str] ->列表[int]
当前回答
Python 2的另一个函数解决方案:
from functools import partial
map(partial(map, int), T1)
Python 3会有一点混乱:
list(map(list, map(partial(map, int), T1)))
我们可以用包装器来解决这个问题
def oldmap(f, iterable):
return list(map(f, iterable))
oldmap(partial(oldmap, int), T1)
其他回答
T3=[]
for i in range(0,len(T1)):
T3.append([])
for j in range(0,len(T1[i])):
b=int(T1[i][j])
T3[i].append(b)
print T3
在Python 3.5.1中，类似这样的工作:
c = input('Enter number:')
print (int(float(c)))
print (round(float(c)))
and
Enter number: 4.7
4
5
看这个函数
def parse_int(s):
try:
res = int(eval(str(s)))
if type(res) == int:
return res
except:
return
Then
val = parse_int('10') # Return 10
val = parse_int('0') # Return 0
val = parse_int('10.5') # Return 10
val = parse_int('0.0') # Return 0
val = parse_int('Ten') # Return None
你也可以检查
if val == None: # True if input value can not be converted
pass # Note: Don't use 'if not val:'
你可以这样做:
T1 = (('13', '17', '18', '21', '32'),
('07', '11', '13', '14', '28'),
('01', '05', '06', '08', '15', '16'))
new_list = list(list(int(a) for a in b if a.isdigit()) for b in T1)
print(new_list)
使用列表推导式:
t2 = [map(int, list(l)) for l in t1]
