blocks|key|1606048|text|int()是Python标准的内置函数，用于将字符串转换为整数值。使用包含数字作为参数的字符串调用它，它将返回转换为整数的数字：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1606049|>>>+int("1")+%2B+1
2|code-block|syntax|javascript|1606050|如果您知道列表T1的结构(它只包含一个级别的列表)，那么您可以在Python3中做到这一点：|1606051|T2+=+[list(map(int,+x))+for+x+in+T1]|1606052|在Python+2中：|1606053|T2+=+[map(int,+x)+for+x+in+T1]|1606054|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/functions.html#int^0|0|5|0|5|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

<a href="https://docs.python.org/3/library/functions.html#int" rel="noreferrer"><code>int()</code></a> 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:
<pre><code>&gt;&gt;&gt; int(&quot;1&quot;) + 1
2
</code></pre>
If you know the structure of your list, T1 (that it simply contains lists, only one level), you could do this in Python 3:
<pre><code>T2 = [list(map(int, x)) for x in T1]
</code></pre>
In Python 2:
<pre><code>T2 = [map(int, x) for x in T1]
</code></pre>

blocks|key|1391240|text|你可以通过列表理解来实现这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391241|T2+=+[[int(column)+for+column+in+row]+for+row+in+T1]|code-block|syntax|javascript|1391242|内部列表理解([int(column)+for+column+in+row])从int中的一系列int-able对象(如十进制字符串)构建一个-able的row。外部列表理解([...+for+row+in+T1]))构建应用于T1中每一项的内部列表理解结果的列表。|offset|length|style|CODE|1391243|如果任何行包含不能由int转换的对象，则代码片段将失败。如果您想处理包含非十进制字符串的行，则需要一个更智能的函数。|1391244|如果您知道行的结构，则可以使用对行的函数的调用来替换内部列表理解。例如：|1391245|T2+=+[parse_a_row_of_T1(row)+for+row+in+T1]|1391246|entityMap^0|0|0|7|V|14|3|1C|3|25|3|2G|K|36|2|0|A|3|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|19|8|@$I|1A|J|1B|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1C|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|T|$]]

You can do this with a list comprehension:

<pre><code>T2 = [[int(column) for column in row] for row in T1]
</code></pre>

The inner list comprehension (<code>[int(column) for column in row]</code>) builds a <code>list</code> of <code>int</code>s from a sequence of <code>int</code>-able objects, like decimal strings, in <code>row</code>. The outer list comprehension (<code>[... for row in T1])</code>) builds a list of the results of the inner list comprehension applied to each item in <code>T1</code>.

The code snippet will fail if any of the rows contain objects that can't be converted by <code>int</code>. 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.

<pre><code>T2 = [parse_a_row_of_T1(row) for row in T1]
</code></pre>

blocks|key|1606254|text|我更喜欢只使用理解列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1606255|[[int(y)+for+y+in+x]+for+x+in+T1]|code-block|syntax|javascript|1606256|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I would rather prefer using only comprehension lists:

<pre><code>[[int(y) for y in x] for x in T1]
</code></pre>

blocks|key|1391484|text|而不是放入int(+)，放入float(+)，这将允许您使用小数和整数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1391485|entityMap^0|5|6|E|8|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Instead of putting <code>int( )</code>, put <code>float( )</code> which will let you use decimals along with integers.

blocks|key|1606187|text|到目前为止，我同意每个人的答案，但问题是，如果你没有所有的整数，它们就会崩溃。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1606188|如果您想排除非整数，那么|1606189|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())|code-block|syntax|javascript|1606190|这只会产生实际的数字。我不使用直接列表理解的原因是因为列表理解会泄漏它们的内部变量。|1606191|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

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

<pre><code>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())
</code></pre>

This yields only actual digits. The reason I don't use direct list comprehensions is because list comprehension leaks their internal variables.

blocks|key|1606435|text|尝尝这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1606436|x+=+"1"|code-block|syntax|javascript|1606437|X是一个字符串，因为它周围有引号，但它里面有一个数字。|1606438|x+=+int(x)|1606439|因为x中有数字1，所以我可以把它变成一个整数。|1606440|要查看字符串是否为数字，可以执行以下操作。|1606441|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)|1606442|它应该打印到IDLE+True，因为x是一个数字。|1606443|y_test+=+is_number(y)

print(y_test)|1606444|它应该打印到IDLE+False，因为y不是数字。|1606445|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|17|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

Try this.

<pre><code>x = "1"
</code></pre>

x is a string because it has quotes around it, but it has a number in it.

<pre><code>x = int(x)
</code></pre>

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.

<pre><code>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)
</code></pre>

It should print to IDLE True because x is a number.

<pre><code>y_test = is_number(y)

print(y_test)
</code></pre>

It should print to IDLE False because y in not a number.

blocks|key|1391407|text|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|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1391408|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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
</code></pre>

blocks|key|1391254|text|使用列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391255|t2+=+[map(int,+list(l))+for+l+in+t1]|code-block|syntax|javascript|1391256|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using list comprehensions:

<pre><code>t2 = [map(int, list(l)) for l in t1]
</code></pre>

blocks|key|1606541|text|Python+2的另一个功能解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1606542|from+functools+import+partial

map(partial(map,+int),+T1)|code-block|syntax|javascript|1606543|不过，Python+3会有一点混乱：|1606544|list(map(list,+map(partial(map,+int),+T1)))|1606545|我们可以用包装器来解决这个问题。|1606546|def+oldmap(f,+iterable):
++++return+list(map(f,+iterable))

oldmap(partial(oldmap,+int),+T1)|1606547|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Yet another functional solution for Python 2:

<pre><code>from functools import partial

map(partial(map, int), T1)
</code></pre>

Python 3 will be a little bit messy though:

<pre><code>list(map(list, map(partial(map, int), T1)))
</code></pre>

we can fix this with a wrapper

<pre><code>def oldmap(f, iterable):
 return list(map(f, iterable))

oldmap(partial(oldmap, int), T1)
</code></pre>

blocks|key|1391699|text|请参阅此函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391700|def+parse_int(s):
++++try:
++++++++res+=+int(eval(str(s)))
++++++++if+type(res)+==+int:
++++++++++++return+res
++++except:
++++++++return|code-block|syntax|javascript|1391701|然后|1391702|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|1391703|您还可以查看|1391704|if+val+==+None:++#+True+if+input+value+can+not+be+converted
++++pass++#+Note:+Don't+use+'if+not+val:'|1391705|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

See this function

<pre><code>def parse_int(s):
 try:
 res = int(eval(str(s)))
 if type(res) == int:
 return res
 except:
 return
</code></pre>

Then

<pre><code>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
</code></pre>

You can also check

<pre><code>if val == None: # True if input value can not be converted
 pass # Note: Don't use 'if not val:'
</code></pre>

blocks|key|1391606|text|在Python+3.5.1中，类似这样的工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391607|c+=+input('Enter+number:')
print+(int(float(c)))
print+(round(float(c)))|code-block|syntax|javascript|1391608|和|1391609|Enter+number:++4.7
4
5|1391610|乔治。|1391611|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In Python 3.5.1 things like these work:

<pre><code>c = input('Enter number:')
print (int(float(c)))
print (round(float(c)))
</code></pre>

and

<pre><code>Enter number: 4.7
4
5
</code></pre>

George.

blocks|key|1391770|text|Python内置了int(string)函数和可选的参数base。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391771|如果您的字符串包含Integer值，它会将其转换为相应的Integer值。但是，如果将十进制数作为字符串，则需要使用float()对其进行转换。|1391772|用法：|1391773|a+=+'22'
b+=+int(a)|code-block|syntax|javascript|1391774|和|1391775|if+a+=+'22.22'
b+=+int(a)+'''will+give+error,+invalid+literal+for+int().'''
b+=+float(a)+'''will+convert+the+string.'''|1391776|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Python has built in function int(string) and optional parameter base.
if your string contains an Integer value, it will convert that to the corresponding Integer value.
However if you have decimnal number as string you'll need float() to convert it.
Usage:
<pre><code>a = '22'
b = int(a)
</code></pre>
and
<pre><code>if a = '22.22'
b = int(a) '''will give error, invalid literal for int().'''
b = float(a) '''will convert the string.'''
</code></pre>

blocks|key|1606589|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1606590|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)++|code-block|syntax|javascript|1606591|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can do something like this:

<pre><code>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) 
</code></pre>

I have a tuple of tuples from a MySQL query like this:

<pre><code>T1 = (('13', '17', '18', '21', '32'),
 ('07', '11', '13', '14', '28'),
 ('01', '05', '06', '08', '15', '16'))
</code></pre>

I'd like to convert all the string elements into integers and put them back into a list of lists:

<pre><code>T2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
</code></pre>

I tried to achieve it with <code>eval</code> but didn't get any decent result yet.

How to convert strings into integers?

Python

MySQL

我有一个来自MySQL查询的元组元组，如下所示：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，并将它们放回列表列表中：T2 = [[13, 17, 18, 21, 32], [7,

问如何将字符串转换为整数？
EN

回答 13

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问如何将字符串转换为整数？
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