问题:如何在Python中将字符串转换为整数?
我有一个来自MySQL查询的元组,像这样:
T1 = (('13', '17', '18', '21', '32'),
('07', '11', '13', '14', '28'),
('01', '05', '06', '08', '15', '16'))
我想将所有字符串元素转换为整数,然后将它们放回列表列表中:
T2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
我试图用它来实现它,eval
但是还没有得到令人满意的结果。
I have a tuple of tuples from a MySQL query like this:
T1 = (('13', '17', '18', '21', '32'),
('07', '11', '13', '14', '28'),
('01', '05', '06', '08', '15', '16'))
I’d like to convert all the string elements into integers and put them back into a list of lists:
T2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
I tried to achieve it with eval
but didn’t get any decent result yet.
回答 0
int()
是Python标准的内置函数,用于将字符串转换为整数值。您使用一个包含数字作为参数的字符串来调用它,它返回转换为整数的数字:
print (int("1") + 1)
上面的照片2
。
如果您知道列表T1的结构(它仅包含列表,仅一个级别),则可以在Python 2中执行此操作:
T2 = [map(int, x) for x in T1]
在Python 3中:
T2 = [list(map(int, x)) for x in T1]
int()
is the Python standard built-in function to convert a string into an integer value. You call it with a string containing a number as the argument, and it returns the number converted to an integer:
print (int("1") + 1)
The above prints 2
.
If you know the structure of your list, T1 (that it simply contains lists, only one level), you could do this in Python 2:
T2 = [map(int, x) for x in T1]
In Python 3:
T2 = [list(map(int, x)) for x in T1]
回答 1
您可以通过列表理解来做到这一点:
T2 = [[int(column) for column in row] for row in T1]
内部列表理解([int(column) for column in row]
)建立一个list
的int
期从序列int
-able物体,如小数字符串中row
。外部列表推导([... for row in T1])
)生成一个内部列表推导的结果的列表,该结果适用于中的每个项目T1
。
如果任何行包含无法通过转换的对象,则代码段将失败int
。如果要处理包含非十进制字符串的行,则需要一个更智能的函数。
如果您知道行的结构,则可以使用对行函数的调用来替换内部列表理解。例如。
T2 = [parse_a_row_of_T1(row) for row in T1]
You can do this with a list comprehension:
T2 = [[int(column) for column in row] for row in T1]
The inner list comprehension ([int(column) for column in row]
) builds a list
of int
s from a sequence of int
-able objects, like decimal strings, in row
. The outer list comprehension ([... for row in T1])
) builds a list of the results of the inner list comprehension applied to each item in T1
.
The code snippet will fail if any of the rows contain objects that can’t be converted by int
. You’ll need a smarter function if you want to process rows containing non-decimal strings.
If you know the structure of the rows, you can replace the inner list comprehension with a call to a function of the row. Eg.
T2 = [parse_a_row_of_T1(row) for row in T1]
回答 2
我宁愿只使用理解列表:
[[int(y) for y in x] for x in T1]
I would rather prefer using only comprehension lists:
[[int(y) for y in x] for x in T1]
回答 3
代替put int( )
,put float( )
可以让您将小数与整数一起使用。
Instead of putting int( )
, put float( )
which will let you use decimals along with integers.
回答 4
到目前为止,我都同意所有人的回答,但是问题是,如果您没有所有整数,它们将崩溃。
如果要排除非整数,则
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) for b in T1 if a.isdigit())
这仅产生实际数字。我不使用直接列表推导的原因是因为列表推导会泄漏其内部变量。
I would agree with everyones answers so far but the problem is is that if you do not have all integers they will crash.
If you wanted to exclude non-integers then
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) for b in T1 if a.isdigit())
This yields only actual digits. The reason I don’t use direct list comprehensions is because list comprehension leaks their internal variables.
回答 5
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
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
回答 6
尝试这个。
x = "1"
x是一个字符串,因为它周围带有引号,但其中带有数字。
x = int(x)
由于x的数字为1,因此我可以将其变成整数。
要查看字符串是否为数字,可以执行此操作。
def is_number(var):
try:
if var == int(var):
return True
except Exception:
return False
x = "1"
y = "test"
x_test = is_number(x)
print(x_test)
它应该打印到IDLE True,因为x是一个数字。
y_test = is_number(y)
print(y_test)
它应该打印为IDLE False,因为y中没有数字。
Try this.
x = "1"
x is a string because it has quotes around it, but it has a number in it.
x = int(x)
Since x has the number 1 in it, I can turn it in to a integer.
To see if a string is a number, you can do this.
def is_number(var):
try:
if var == int(var):
return True
except Exception:
return False
x = "1"
y = "test"
x_test = is_number(x)
print(x_test)
It should print to IDLE True because x is a number.
y_test = is_number(y)
print(y_test)
It should print to IDLE False because y in not a number.
回答 7
使用列表推导:
t2 = [map(int, list(l)) for l in t1]
Using list comprehensions:
t2 = [map(int, list(l)) for l in t1]
回答 8
在Python 3.5.1中,这些工作如下:
c = input('Enter number:')
print (int(float(c)))
print (round(float(c)))
和
Enter number: 4.7
4
5
乔治。
In Python 3.5.1 things like these work:
c = input('Enter number:')
print (int(float(c)))
print (round(float(c)))
and
Enter number: 4.7
4
5
George.
回答 9
查看此功能
def parse_int(s):
try:
res = int(eval(str(s)))
if type(res) == int:
return res
except:
return
然后
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
您也可以检查
if val == None: # True if input value can not be converted
pass # Note: Don't use 'if not val:'
See this function
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
You can also check
if val == None: # True if input value can not be converted
pass # Note: Don't use 'if not val:'
回答 10
适用于Python 2的另一个功能解决方案:
from functools import partial
map(partial(map, int), T1)
不过,Python 3会有些混乱:
list(map(list, map(partial(map, int), T1)))
我们可以用包装纸解决
def oldmap(f, iterable):
return list(map(f, iterable))
oldmap(partial(oldmap, int), T1)
Yet another functional solution for Python 2:
from functools import partial
map(partial(map, int), T1)
Python 3 will be a little bit messy though:
list(map(list, map(partial(map, int), T1)))
we can fix this with a wrapper
def oldmap(f, iterable):
return list(map(f, iterable))
oldmap(partial(oldmap, int), T1)
回答 11
如果只是元组的元组,类似 rows=[map(int, row) for row in rows]
就可以解决。(在其中有一个列表推导和对map(f,lst)的调用,该调用等于[f in a lst]中的f(a)。)
如果由于某种原因在数据库中有类似的东西,Eval 不是您想要做的__import__("os").unlink("importantsystemfile")
。始终验证您的输入(如果没有其他问题,如果输入错误,则会引发int()异常)。
If it’s only a tuple of tuples, something like rows=[map(int, row) for row in rows]
will do the trick. (There’s a list comprehension and a call to map(f, lst), which is equal to [f(a) for a in lst], in there.)
Eval is not what you want to do, in case there’s something like __import__("os").unlink("importantsystemfile")
in your database for some reason.
Always validate your input (if with nothing else, the exception int() will raise if you have bad input).
回答 12
您可以执行以下操作:
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)
You can do something like this:
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)
回答 13
我想分享一个似乎此处未提及的可用选项:
rumpy.random.permutation(x)
将生成数组x的随机排列。不完全是您的要求,但这是解决类似问题的潜在方法。
I want to share an available option that doesn’t seem to be mentioned here yet:
rumpy.random.permutation(x)
Will generate a random permutation of array x. Not exactly what you asked for, but it is a potential solution to similar questions.
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