Python 实用宝典

Question 1

I think the maximum integer in python is available by calling sys.maxint.

What is the maximum float or long in Python?

Question 2

For float have a look at sys.float_info:

>>> import sys
>>> sys.float_info
sys.floatinfo(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308, min=2.2
250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15, mant_dig=53, epsil
on=2.2204460492503131e-16, radix=2, rounds=1)

Specifically, sys.float_info.max:

>>> sys.float_info.max
1.7976931348623157e+308

If that’s not big enough, there’s always positive infinity:

>>> infinity = float("inf")
>>> infinity
inf
>>> infinity / 10000
inf

The long type has unlimited precision, so I think you’re only limited by available memory.

Question 3

sys.maxint is not the largest integer supported by python. It’s the largest integer supported by python’s regular integer type.

Question 4

If you are using numpy, you can use dtype ‘float128‘ and get a max float of 10e+4931

>>> np.finfo(np.float128)
finfo(resolution=1e-18, min=-1.18973149536e+4932, max=1.18973149536e+4932, dtype=float128)

Question 5

Is there a way to convert true of type unicode to 1 and false of type unicode to 0 (in Python)?

For example: x == 'true' and type(x) == unicode

I want x = 1

PS: I don’t want to use if–else.

Question 6

Use int() on a boolean test:

x = int(x == 'true')

int() turns the boolean into 1 or 0. Note that any value not equal to 'true' will result in 0 being returned.

Question 7

If B is a Boolean array, write

B = B*1

(A bit code golfy.)

Question 8

You can use x.astype('uint8') where x is your Boolean array.

Question 9

Here’s a yet another solution to your problem:

def to_bool(s):
    return 1 - sum(map(ord, s)) % 2
    # return 1 - sum(s.encode('ascii')) % 2  # Alternative for Python 3

It works because the sum of the ASCII codes of 'true' is 448, which is even, while the sum of the ASCII codes of 'false' is 523 which is odd.

The funny thing about this solution is that its result is pretty random if the input is not one of 'true' or 'false'. Half of the time it will return 0, and the other half 1. The variant using encode will raise an encoding error if the input is not ASCII (thus increasing the undefined-ness of the behaviour).

Seriously, I believe the most readable, and faster, solution is to use an if:

def to_bool(s):
    return 1 if s == 'true' else 0

See some microbenchmarks:

In [14]: def most_readable(s):
    ...:     return 1 if s == 'true' else 0

In [15]: def int_cast(s):
    ...:     return int(s == 'true')

In [16]: def str2bool(s):
    ...:     try:
    ...:         return ['false', 'true'].index(s)
    ...:     except (ValueError, AttributeError):
    ...:         raise ValueError()

In [17]: def str2bool2(s):
    ...:     try:
    ...:         return ('false', 'true').index(s)
    ...:     except (ValueError, AttributeError):
    ...:         raise ValueError()

In [18]: def to_bool(s):
    ...:     return 1 - sum(s.encode('ascii')) % 2

In [19]: %timeit most_readable('true')
10000000 loops, best of 3: 112 ns per loop

In [20]: %timeit most_readable('false')
10000000 loops, best of 3: 109 ns per loop

In [21]: %timeit int_cast('true')
1000000 loops, best of 3: 259 ns per loop

In [22]: %timeit int_cast('false')
1000000 loops, best of 3: 262 ns per loop

In [23]: %timeit str2bool('true')
1000000 loops, best of 3: 343 ns per loop

In [24]: %timeit str2bool('false')
1000000 loops, best of 3: 325 ns per loop

In [25]: %timeit str2bool2('true')
1000000 loops, best of 3: 295 ns per loop

In [26]: %timeit str2bool2('false')
1000000 loops, best of 3: 277 ns per loop

In [27]: %timeit to_bool('true')
1000000 loops, best of 3: 607 ns per loop

In [28]: %timeit to_bool('false')
1000000 loops, best of 3: 612 ns per loop

Notice how the if solution is at least 2.5x times faster than all the other solutions. It does not make sense to put as a requirement to avoid using ifs except if this is some kind of homework (in which case you shouldn’t have asked this in the first place).

Question 10

If you need a general purpose conversion from a string which per se is not a bool, you should better write a routine similar to the one depicted below. In keeping with the spirit of duck typing, I have not silently passed the error but converted it as appropriate for the current scenario.

>>> def str2bool(st):
try:
    return ['false', 'true'].index(st.lower())
except (ValueError, AttributeError):
    raise ValueError('no Valid Conversion Possible')


>>> str2bool('garbaze')

Traceback (most recent call last):
  File "<pyshell#106>", line 1, in <module>
    str2bool('garbaze')
  File "<pyshell#105>", line 5, in str2bool
    raise TypeError('no Valid COnversion Possible')
TypeError: no Valid Conversion Possible
>>> str2bool('false')
0
>>> str2bool('True')
1

Question 11

bool to int: x = (x == 'true') + 0

Now the x contains 1 if x == 'true' else 0.

Note: x == 'true' will return bool which then will be typecasted to int having value (1 if bool value is True else 0) when added with 0.

Question 12

only with this:

const a = true; const b = false;

console.log(+a);//1 console.log(+b);//0

Question 13

在python 2.7.6中使用新的Enum功能（通过backport enum34）。

给定以下定义，如何将int转换为相应的Enum值？

from enum import Enum

class Fruit(Enum):
    Apple = 4
    Orange = 5
    Pear = 6

我知道我可以手工制作一系列的if语句来进行转换，但是有没有简单的pythonic转换方法？基本上，我想要一个返回枚举值的函数ConvertIntToFruit（int）。

我的用例是我有一个记录的csv文件，在其中我将每个记录读入一个对象。文件字段之一是代表枚举的整数字段。在填充对象时，我想将文件中的整数字段转换为对象中对应的Enum值。

Question 14

Using the new Enum feature (via backport enum34) with python 2.7.6.

Given the following definition, how can I convert an int to the corresponding Enum value?

from enum import Enum

class Fruit(Enum):
    Apple = 4
    Orange = 5
    Pear = 6

I know I can hand craft a series of if-statements to do the conversion but is there an easy pythonic way to convert? Basically, I’d like a function ConvertIntToFruit(int) that returns an enum value.

My use case is I have a csv file of records where I’m reading each record into an object. One of the file fields is an integer field that represents an enumeration. As I’m populating the object I’d like to convert that integer field from the file into the corresponding Enum value in the object.

Question 15

您“打电话”Enum上课：

Fruit(5)

轮到5为Fruit.Orange：

>>> from enum import Enum
>>> class Fruit(Enum):
...     Apple = 4
...     Orange = 5
...     Pear = 6
... 
>>> Fruit(5)
<Fruit.Orange: 5>

从文档的程序访问到枚举成员及其属性部分：

有时，以编程方式访问枚举中的成员很有用（例如，Color.red由于在编写程序时尚不知道确切的颜色而无法这样做）。Enum允许这样的访问：
>>> Color(1)
<Color.red: 1>
>>> Color(3)
<Color.blue: 3>

在相关说明中：要映射包含枚举成员名称的字符串值，请使用subscription：

>>> s = 'Apple'
>>> Fruit[s]
<Fruit.Apple: 4>

Question 16

You ‘call’ the Enum class:

Fruit(5)

to turn 5 into Fruit.Orange:

>>> from enum import Enum
>>> class Fruit(Enum):
...     Apple = 4
...     Orange = 5
...     Pear = 6
... 
>>> Fruit(5)
<Fruit.Orange: 5>

From the Programmatic access to enumeration members and their attributes section of the documentation:

Sometimes it’s useful to access members in enumerations programmatically (i.e. situations where Color.red won’t do because the exact color is not known at program-writing time). Enum allows such access:
>>> Color(1)
<Color.red: 1>
>>> Color(3)
<Color.blue: 3>

In a related note: to map a string value containing the name of an enum member, use subscription:

>>> s = 'Apple'
>>> Fruit[s]
<Fruit.Apple: 4>

Question 17

我认为这是简单的话是对转换int价值为Enum通过调用EnumType(int_value)，访问后name的的Enum对象：

my_fruit_from_int = Fruit(5) #convert to int
fruit_name = my_fruit_from_int.name #get the name
print(fruit_name) #Orange will be printed here

或作为功能：

def convert_int_to_fruit(int_value):
    try:
        my_fruit_from_int = Fruit(int_value)
        return my_fruit_from_int.name
    except:
        return None

Question 18

I think it is in simple words is to convert the int value into Enum by calling EnumType(int_value), after that access the name of the Enum object:

my_fruit_from_int = Fruit(5) #convert to int
fruit_name = my_fruit_from_int.name #get the name
print(fruit_name) #Orange will be printed here

Or as a function:

def convert_int_to_fruit(int_value):
    try:
        my_fruit_from_int = Fruit(int_value)
        return my_fruit_from_int.name
    except:
        return None

Question 19

我想要类似的东西，以便可以从单个引用访问值对的任何一部分。香草版本：

#!/usr/bin/env python3


from enum import IntEnum


class EnumDemo(IntEnum):
    ENUM_ZERO       = 0
    ENUM_ONE        = 1
    ENUM_TWO        = 2
    ENUM_THREE      = 3
    ENUM_INVALID    = 4


#endclass.


print('Passes')
print('1) %d'%(EnumDemo['ENUM_TWO']))
print('2) %s'%(EnumDemo['ENUM_TWO']))
print('3) %s'%(EnumDemo.ENUM_TWO.name))
print('4) %d'%(EnumDemo.ENUM_TWO))
print()


print('Fails')
print('1) %d'%(EnumDemo.ENUM_TWOa))

失败将引发异常。

一个更强大的版本：

#!/usr/bin/env python3


class EnumDemo():


    enumeration =   (
                        'ENUM_ZERO',    # 0.
                        'ENUM_ONE',     # 1.
                        'ENUM_TWO',     # 2.
                        'ENUM_THREE',   # 3.
                        'ENUM_INVALID'  # 4.
                    )


    def name(self, val):
        try:

            name = self.enumeration[val]
        except IndexError:

            # Always return last tuple.
            name = self.enumeration[len(self.enumeration) - 1]

        return name


    def number(self, val):
        try:

            index = self.enumeration.index(val)
        except (TypeError, ValueError):

            # Always return last tuple.
            index = (len(self.enumeration) - 1)

        return index


#endclass.


print('Passes')
print('1) %d'%(EnumDemo().number('ENUM_TWO')))
print('2) %s'%(EnumDemo().number('ENUM_TWO')))
print('3) %s'%(EnumDemo().name(1)))
print('4) %s'%(EnumDemo().enumeration[1]))
print()
print('Fails')
print('1) %d'%(EnumDemo().number('ENUM_THREEa')))
print('2) %s'%(EnumDemo().number('ENUM_THREEa')))
print('3) %s'%(EnumDemo().name(11)))
print('4) %s'%(EnumDemo().enumeration[-1]))

如果使用不正确，这可以避免产生异常，而是传回故障指示。一种更Python化的方法是返回“ None”，但是我的特定应用程序直接使用文本。

Question 20

I wanted something similar so that I could access either part of the value pair from a single reference. The vanilla version:

#!/usr/bin/env python3


from enum import IntEnum


class EnumDemo(IntEnum):
    ENUM_ZERO       = 0
    ENUM_ONE        = 1
    ENUM_TWO        = 2
    ENUM_THREE      = 3
    ENUM_INVALID    = 4


#endclass.


print('Passes')
print('1) %d'%(EnumDemo['ENUM_TWO']))
print('2) %s'%(EnumDemo['ENUM_TWO']))
print('3) %s'%(EnumDemo.ENUM_TWO.name))
print('4) %d'%(EnumDemo.ENUM_TWO))
print()


print('Fails')
print('1) %d'%(EnumDemo.ENUM_TWOa))

The failure throws an exception as would be expected.

A more robust version:

#!/usr/bin/env python3


class EnumDemo():


    enumeration =   (
                        'ENUM_ZERO',    # 0.
                        'ENUM_ONE',     # 1.
                        'ENUM_TWO',     # 2.
                        'ENUM_THREE',   # 3.
                        'ENUM_INVALID'  # 4.
                    )


    def name(self, val):
        try:

            name = self.enumeration[val]
        except IndexError:

            # Always return last tuple.
            name = self.enumeration[len(self.enumeration) - 1]

        return name


    def number(self, val):
        try:

            index = self.enumeration.index(val)
        except (TypeError, ValueError):

            # Always return last tuple.
            index = (len(self.enumeration) - 1)

        return index


#endclass.


print('Passes')
print('1) %d'%(EnumDemo().number('ENUM_TWO')))
print('2) %s'%(EnumDemo().number('ENUM_TWO')))
print('3) %s'%(EnumDemo().name(1)))
print('4) %s'%(EnumDemo().enumeration[1]))
print()
print('Fails')
print('1) %d'%(EnumDemo().number('ENUM_THREEa')))
print('2) %s'%(EnumDemo().number('ENUM_THREEa')))
print('3) %s'%(EnumDemo().name(11)))
print('4) %s'%(EnumDemo().enumeration[-1]))

When not used correctly this avoids creating an exception and, instead, passes back a fault indication. A more Pythonic way to do this would be to pass back “None” but my particular application uses the text directly.

Question 21

I have a dataframe in pandas with mixed int and str data columns. I want to concatenate first the columns within the dataframe. To do that I have to convert an int column to str. I’ve tried to do as follows:

mtrx['X.3'] = mtrx.to_string(columns = ['X.3'])

or

mtrx['X.3'] = mtrx['X.3'].astype(str)

but in both cases it’s not working and I’m getting an error saying “cannot concatenate ‘str’ and ‘int’ objects”. Concatenating two str columns is working perfectly fine.

Question 22

In [16]: df = DataFrame(np.arange(10).reshape(5,2),columns=list('AB'))

In [17]: df
Out[17]: 
   A  B
0  0  1
1  2  3
2  4  5
3  6  7
4  8  9

In [18]: df.dtypes
Out[18]: 
A    int64
B    int64
dtype: object

Convert a series

In [19]: df['A'].apply(str)
Out[19]: 
0    0
1    2
2    4
3    6
4    8
Name: A, dtype: object

In [20]: df['A'].apply(str)[0]
Out[20]: '0'

Don’t forget to assign the result back:

df['A'] = df['A'].apply(str)

Convert the whole frame

In [21]: df.applymap(str)
Out[21]: 
   A  B
0  0  1
1  2  3
2  4  5
3  6  7
4  8  9

In [22]: df.applymap(str).iloc[0,0]
Out[22]: '0'

df = df.applymap(str)

Question 23

Change data type of DataFrame column:

To int:

df.column_name = df.column_name.astype(np.int64)

To str:

df.column_name = df.column_name.astype(str)

Question 24

Warning: Both solutions given ( astype() and apply() ) do not preserve NULL values in either the nan or the None form.

import pandas as pd
import numpy as np

df = pd.DataFrame([None,'string',np.nan,42], index=[0,1,2,3], columns=['A'])

df1 = df['A'].astype(str)
df2 =  df['A'].apply(str)

print df.isnull()
print df1.isnull()
print df2.isnull()

I believe this is fixed by the implementation of to_string()

Question 25

Use the following code:

df.column_name = df.column_name.astype('str')

Question 26

Just for an additional reference.

All of the above answers will work in case of a data frame. But if you are using lambda while creating / modify a column this won’t work, Because there it is considered as a int attribute instead of pandas series. You have to use str( target_attribute ) to make it as a string. Please refer the below example.

def add_zero_in_prefix(df):
    if(df['Hour']<10):
        return '0' + str(df['Hour'])

data['str_hr'] = data.apply(add_zero_in_prefix, axis=1)

Question 27

Is there a preferred way to keep the data type of a numpy array fixed as int (or int64 or whatever), while still having an element inside listed as numpy.NaN?

In particular, I am converting an in-house data structure to a Pandas DataFrame. In our structure, we have integer-type columns that still have NaN’s (but the dtype of the column is int). It seems to recast everything as a float if we make this a DataFrame, but we’d really like to be int.

Thoughts?

Things tried:

I tried using the from_records() function under pandas.DataFrame, with coerce_float=False and this did not help. I also tried using NumPy masked arrays, with NaN fill_value, which also did not work. All of these caused the column data type to become a float.

Question 28

This capability has been added to pandas (beginning with version 0.24): https://pandas.pydata.org/pandas-docs/version/0.24/whatsnew/v0.24.0.html#optional-integer-na-support

At this point, it requires the use of extension dtype Int64 (capitalized), rather than the default dtype int64 (lowercase).

Question 29

NaN can’t be stored in an integer array. This is a known limitation of pandas at the moment; I have been waiting for progress to be made with NA values in NumPy (similar to NAs in R), but it will be at least 6 months to a year before NumPy gets these features, it seems:

http://pandas.pydata.org/pandas-docs/stable/gotchas.html#support-for-integer-na

(This feature has been added beginning with version 0.24 of pandas, but note it requires the use of extension dtype Int64 (capitalized), rather than the default dtype int64 (lower case): https://pandas.pydata.org/pandas-docs/version/0.24/whatsnew/v0.24.0.html#optional-integer-na-support )

Question 30

If performance is not the main issue, you can store strings instead.

df.col = df.col.dropna().apply(lambda x: str(int(x)) )

Then you can mix then with NaN as much as you want. If you really want to have integers, depending on your application, you can use -1, or 0, or 1234567890, or some other dedicated value to represent NaN.

You can also temporarily duplicate the columns: one as you have, with floats; the other one experimental, with ints or strings. Then inserts asserts in every reasonable place checking that the two are in sync. After enough testing you can let go of the floats.

Question 31

This is not a solution for all cases, but mine (genomic coordinates) I’ve resorted to using 0 as NaN

a3['MapInfo'] = a3['MapInfo'].fillna(0).astype(int)

This at least allows for the proper ‘native’ column type to be used, operations like subtraction, comparison etc work as expected

Question 32

Pandas v0.24+

Functionality to support NaN in integer series will be available in v0.24 upwards. There’s information on this in the v0.24 “What’s New” section, and more details under Nullable Integer Data Type.

Pandas v0.23 and earlier

In general, it’s best to work with float series where possible, even when the series is upcast from int to float due to inclusion of NaN values. This enables vectorised NumPy-based calculations where, otherwise, Python-level loops would be processed.

The docs do suggest : “One possibility is to use dtype=object arrays instead.” For example:

s = pd.Series([1, 2, 3, np.nan])

print(s.astype(object))

0      1
1      2
2      3
3    NaN
dtype: object

For cosmetic reasons, e.g. output to a file, this may be preferable.

Pandas v0.23 and earlier: background

NaN is considered a float. The docs currently (as of v0.23) specify the reason why integer series are upcasted to float:

In the absence of high performance NA support being built into NumPy from the ground up, the primary casualty is the ability to represent NAs in integer arrays.

This trade-off is made largely for memory and performance reasons, and also so that the resulting Series continues to be “numeric”.

The docs also provide rules for upcasting due to NaN inclusion:

Typeclass   Promotion dtype for storing NAs
floating    no change
object      no change
integer     cast to float64
boolean     cast to object

Question 33

This is now possible, since pandas v 0.24.0

pandas 0.24.x release notes Quote: “Pandas has gained the ability to hold integer dtypes with missing values.

Question 34

Just wanted to add that in case you are trying to convert a float (1.143) vector to integer (1) that has NA converting to the new ‘Int64’ dtype will give you an error. In order to solve this you have to round the numbers and then do “.astype(‘Int64’)”

s1 = pd.Series([1.434, 2.343, np.nan])
#without round() the next line returns an error 
s1.astype('Int64')
#cannot safely cast non-equivalent float64 to int64
##with round() it works
s1.round().astype('Int64')
0      1
1      2
2    NaN
dtype: Int64

My use case is that I have a float series that I want to round to int, but when you do .round() a ‘*.0’ at the end of the number remains, so you can drop that 0 from the end by converting to int.

Question 35

If there are blanks in the text data, columns that would normally be integers will be cast to floats as float64 dtype because int64 dtype cannot handle nulls. This can cause inconsistent schema if you are loading multiple files some with blanks (which will end up as float64 and others without which will end up as int64

This code will attempt to convert any number type columns to Int64 (as opposed to int64) since Int64 can handle nulls

import pandas as pd
import numpy as np

#show datatypes before transformation
mydf.dtypes

for c in mydf.select_dtypes(np.number).columns:
    try:
        mydf[c] = mydf[c].astype('Int64')
        print('casted {} as Int64'.format(c))
    except:
        print('could not cast {} to Int64'.format(c))

#show datatypes after transformation
mydf.dtypes

Question 36

I want to take an integer (that will be <= 255), to a hex string representation

e.g.: I want to pass in 65 and get out '\x41', or 255 and get '\xff'.

I’ve tried doing this with the struct.pack('c',65), but that chokes on anything above 9 since it wants to take in a single character string.

Question 37

You are looking for the chr function.

You seem to be mixing decimal representations of integers and hex representations of integers, so it’s not entirely clear what you need. Based on the description you gave, I think one of these snippets shows what you want.

>>> chr(0x65) == '\x65'
True


>>> hex(65)
'0x41'
>>> chr(65) == '\x41'
True

Note that this is quite different from a string containing an integer as hex. If that is what you want, use the hex builtin.

Question 38

This will convert an integer to a 2 digit hex string with the 0x prefix:

strHex = "0x%0.2X" % 255

Question 39

What about hex()?

hex(255)  # 0xff

If you really want to have \ in front you can do:

print '\\' + hex(255)[1:]

Question 40

Try:

"0x%x" % 255 # => 0xff

or

"0x%X" % 255 # => 0xFF

Python Documentation says: “keep this under Your pillow: http://docs.python.org/library/index.html“

Question 41

Let me add this one, because sometimes you just want the single digit representation

( x can be lower, ‘x’, or uppercase, ‘X’, the choice determines if the output letters are upper or lower.):

'{:x}'.format(15)
> f

And now with the new f'' format strings you can do:

f'{15:x}'
> f

To add 0 padding you can use 0>n:

f'{2034:0>4X}'
> 07F2

NOTE: the initial ‘f’ in f'{15:x}' is to signify a format string

Question 42

If you want to pack a struct with a value <255 (one byte unsigned, uint8_t) and end up with a string of one character, you’re probably looking for the format B instead of c. C converts a character to a string (not too useful by itself) while B converts an integer.

struct.pack('B', 65)

(And yes, 65 is \x41, not \x65.)

The struct class will also conveniently handle endianness for communication or other uses.

Question 43

Note that for large values, hex() still works (some other answers don’t):

x = hex(349593196107334030177678842158399357)
print(x)

Python 2: 0x4354467b746f6f5f736d616c6c3f7dL
Python 3: 0x4354467b746f6f5f736d616c6c3f7d

For a decrypted RSA message, one could do the following:

import binascii

hexadecimals = hex(349593196107334030177678842158399357)

print(binascii.unhexlify(hexadecimals[2:-1])) # python 2
print(binascii.unhexlify(hexadecimals[2:])) # python 3

Question 44

This worked best for me

"0x%02X" % 5  # => 0x05
"0x%02X" % 17 # => 0x11

Change the (2) if you want a number with a bigger width (2 is for 2 hex printned chars) so 3 will give you the following

"0x%03X" % 5  # => 0x005
"0x%03X" % 17 # => 0x011

Question 45

I wanted a random integer converted into a six-digit hex string with a # at the beginning. To get this I used

"#%6x" % random.randint(0xFFFFFF)

Question 46

With format(), as per format-examples, we can do:

>>> # format also supports binary numbers
>>> "int: {0:d};  hex: {0:x};  oct: {0:o};  bin: {0:b}".format(42)
'int: 42;  hex: 2a;  oct: 52;  bin: 101010'
>>> # with 0x, 0o, or 0b as prefix:
>>> "int: {0:d};  hex: {0:#x};  oct: {0:#o};  bin: {0:#b}".format(42)
'int: 42;  hex: 0x2a;  oct: 0o52;  bin: 0b101010'

Question 47

(int_variable).to_bytes(bytes_length, byteorder='big'|'little').hex()

For example:

>>> (434).to_bytes(4, byteorder='big').hex()
'000001b2'
>>> (434).to_bytes(4, byteorder='little').hex()
'b2010000'

Question 48

Also you can convert any number in any base to hex. Use this one line code here it’s easy and simple to use:

hex(int(n,x)).replace("0x","")

You have a string n that is your number and x the base of that number. First, change it to integer and then to hex but hex has 0x at the first of it so with replace we remove it.

Question 49

As an alternative representation you could use

[in] '%s' % hex(15)
[out]'0xf'

Question 50

Why are x and y strings instead of ints in the below code?

(Note: in Python 2.x use raw_input(). In Python 3.x use input(). raw_input() was renamed to input() in Python 3.x)

play = True

while play:

    x = input("Enter a number: ")
    y = input("Enter a number: ")

    print(x + y)
    print(x - y)
    print(x * y)
    print(x / y)
    print(x % y)

    if input("Play again? ") == "no":
        play = False

Question 51

TLDR

Python 3 doesn’t evaluate the data received with input function, but Python 2’s input function does (read the next section to understand the implication).
Python 2’s equivalent of Python 3’s input is the raw_input function.

Python 2.x

There were two functions to get user input, called input and raw_input. The difference between them is, raw_input doesn’t evaluate the data and returns as it is, in string form. But, input will evaluate whatever you entered and the result of evaluation will be returned. For example,

>>> import sys
>>> sys.version
'2.7.6 (default, Mar 22 2014, 22:59:56) \n[GCC 4.8.2]'
>>> data = input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
(22, <type 'int'>)

The data 5 + 17 is evaluated and the result is 22. When it evaluates the expression 5 + 17, it detects that you are adding two numbers and so the result will also be of the same int type. So, the type conversion is done for free and 22 is returned as the result of input and stored in data variable. You can think of input as the raw_input composed with an eval call.

>>> data = eval(raw_input("Enter a number: "))
Enter a number: 5 + 17
>>> data, type(data)
(22, <type 'int'>)

Note: you should be careful when you are using input in Python 2.x. I explained why one should be careful when using it, in this answer.

But, raw_input doesn’t evaluate the input and returns as it is, as a string.

>>> import sys
>>> sys.version
'2.7.6 (default, Mar 22 2014, 22:59:56) \n[GCC 4.8.2]'
>>> data = raw_input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
('5 + 17', <type 'str'>)

Python 3.x

Python 3.x’s input and Python 2.x’s raw_input are similar and raw_input is not available in Python 3.x.

>>> import sys
>>> sys.version
'3.4.0 (default, Apr 11 2014, 13:05:11) \n[GCC 4.8.2]'
>>> data = input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
('5 + 17', <class 'str'>)

Solution

To answer your question, since Python 3.x doesn’t evaluate and convert the data type, you have to explicitly convert to ints, with int, like this

x = int(input("Enter a number: "))
y = int(input("Enter a number: "))

You can accept numbers of any base and convert them directly to base-10 with the int function, like this

>>> data = int(input("Enter a number: "), 8)
Enter a number: 777
>>> data
511
>>> data = int(input("Enter a number: "), 16)
Enter a number: FFFF
>>> data
65535
>>> data = int(input("Enter a number: "), 2)
Enter a number: 10101010101
>>> data
1365

The second parameter tells what is the base of the numbers entered and then internally it understands and converts it. If the entered data is wrong it will throw a ValueError.

>>> data = int(input("Enter a number: "), 2)
Enter a number: 1234
Traceback (most recent call last):
  File "<input>", line 1, in <module>
ValueError: invalid literal for int() with base 2: '1234'

For values that can have a fractional component, the type would be float rather than int:

x = float(input("Enter a number:"))

Apart from that, your program can be changed a little bit, like this

while True:
    ...
    ...
    if input("Play again? ") == "no":
        break

You can get rid of the play variable by using break and while True.

Question 52

In Python 3.x, raw_input was renamed to input and the Python 2.x input was removed.

This means that, just like raw_input, input in Python 3.x always returns a string object.

To fix the problem, you need to explicitly make those inputs into integers by putting them in int:

x = int(input("Enter a number: "))
y = int(input("Enter a number: "))

Question 53

For multiple integer in a single line, map might be better.

arr = map(int, raw_input().split())

If the number is already known, (like 2 integers), you can use

num1, num2 = map(int, raw_input().split())

Question 54

input() (Python 3) and raw_input() (Python 2) always return strings. Convert the result to integer explicitly with int().

x = int(input("Enter a number: "))
y = int(input("Enter a number: "))

Question 55

Multiple questions require input for several integers on single line. The best way is to input the whole string of numbers one one line and then split them to integers. Here is a Python 3 version:

a = []
p = input()
p = p.split()      
for i in p:
    a.append(int(i))

Also a list comprehension can be used

p = input().split("whatever the seperator is")

And to convert all the inputs from string to int we do the following

x = [int(i) for i in p]
print(x, end=' ')

shall print the list elements in a straight line.

Question 56

Convert to integers:

my_number = int(input("enter the number"))

Similarly for floating point numbers:

my_decimalnumber = float(input("enter the number"))

Question 57

n=int(input())
for i in range(n):
    n=input()
    n=int(n)
    arr1=list(map(int,input().split()))

the for loop shall run ‘n’ number of times . the second ‘n’ is the length of the array. the last statement maps the integers to a list and takes input in space separated form . you can also return the array at the end of for loop.

Question 58

I encountered a problem of taking integer input while solving a problem on CodeChef, where two integers – separated by space – should be read from one line.

While int(input()) is sufficient for a single integer, I did not find a direct way to input two integers. I tried this:

num = input()
num1 = 0
num2 = 0

for i in range(len(num)):
    if num[i] == ' ':
        break

num1 = int(num[:i])
num2 = int(num[i+1:])

Now I use num1 and num2 as integers. Hope this helps.

Question 59

def dbz():
    try:
        r = raw_input("Enter number:")
        if r.isdigit():
            i = int(raw_input("Enter divident:"))
            d = int(r)/i
            print "O/p is -:",d
        else:
            print "Not a number"
    except Exception ,e:
        print "Program halted incorrect data entered",type(e)
dbz()

Or 

num = input("Enter Number:")#"input" will accept only numbers

Question 60

While in your example, int(input(...)) does the trick in any case, python-future‘s builtins.input is worth consideration since that makes sure your code works for both Python 2 and 3 and disables Python2’s default behaviour of input trying to be “clever” about the input data type (builtins.input basically just behaves like raw_input).

问题：Python中的最大浮点数是多少？

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问题：NumPy或Pandas：具有NaN值时，将数组类型保持为整数

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熊猫v0.24 +

Pandas v0.23及更早版本

熊猫v0.23及更早版本：背景

Pandas v0.24+

Pandas v0.23 and earlier

Pandas v0.23 and earlier: background

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问题：“ is”运算符对整数的行为异常

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Python的“ is”运算符在使用整数时表现异常吗？

说明

问题

是什么is呢

那我们为什么要使用is呢？

Python’s “is” operator behaves unexpectedly with integers?

Explanation

The Question

What is does

Why would we want to use is then?

回答 2

是什么`is`呢

那我们为什么要使用`is`呢？

What `is` does

Why would we want to use `is` then?

但是，`257 is 257`？这是怎么回事？

But, `257 is 257`? What’s up?