Convert values in column from hex to binary in pandas data frame

I have one column in pandas data frame with hex values, for example:

Data

1A

2B

BB

FF

A7

78

CB

I want to convert hex values in binary, then from binary to take first 3 bits and finally convert 3 bits value in decimal.

Data column in binary will be:

the first 3 bits:

and finally the desired value in decimal:

How to do this? I tried with bin() function, but it doesn't work with pandas data frames.

asked Nov 11 at 21:06

slobokv83

729

1

Can you share your work so that it ll be easy for us to replicate.
– Naveen
Nov 11 at 21:12

I shared everything I could...I wish I could share more, but it is not allowed.
– slobokv83
Nov 12 at 8:11

add a comment |

I have one column in pandas data frame with hex values, for example:

Data

1A

2B

BB

FF

A7

78

CB

I want to convert hex values in binary, then from binary to take first 3 bits and finally convert 3 bits value in decimal.

Data column in binary will be:

the first 3 bits:

and finally the desired value in decimal:

How to do this? I tried with bin() function, but it doesn't work with pandas data frames.

asked Nov 11 at 21:06

slobokv83

729

1

Can you share your work so that it ll be easy for us to replicate.
– Naveen
Nov 11 at 21:12

I shared everything I could...I wish I could share more, but it is not allowed.
– slobokv83
Nov 12 at 8:11

add a comment |

I have one column in pandas data frame with hex values, for example:

Data

1A

2B

BB

FF

A7

78

CB

I want to convert hex values in binary, then from binary to take first 3 bits and finally convert 3 bits value in decimal.

Data column in binary will be:

the first 3 bits:

and finally the desired value in decimal:

How to do this? I tried with bin() function, but it doesn't work with pandas data frames.

asked Nov 11 at 21:06

slobokv83

729

I have one column in pandas data frame with hex values, for example:

Data

1A

2B

BB

FF

A7

78

CB

I want to convert hex values in binary, then from binary to take first 3 bits and finally convert 3 bits value in decimal.

Data column in binary will be:

the first 3 bits:

and finally the desired value in decimal:

How to do this? I tried with bin() function, but it doesn't work with pandas data frames.

python pandas dataframe binary hex

asked Nov 11 at 21:06

slobokv83

729

asked Nov 11 at 21:06

slobokv83

729

asked Nov 11 at 21:06

slobokv83

729

asked Nov 11 at 21:06

slobokv83

729

asked Nov 11 at 21:06

slobokv83

729

1

Can you share your work so that it ll be easy for us to replicate.
– Naveen
Nov 11 at 21:12

I shared everything I could...I wish I could share more, but it is not allowed.
– slobokv83
Nov 12 at 8:11

add a comment |

1

Can you share your work so that it ll be easy for us to replicate.
– Naveen
Nov 11 at 21:12

I shared everything I could...I wish I could share more, but it is not allowed.
– slobokv83
Nov 12 at 8:11

Can you share your work so that it ll be easy for us to replicate.
– Naveen
Nov 11 at 21:12

I shared everything I could...I wish I could share more, but it is not allowed.
– slobokv83
Nov 12 at 8:11

add a comment |

2 Answers
2

active

oldest

votes

We can do this by a chain of actions:

first we convert the hexadecimal number to an int with .apply(int, base=16);

next we convert this to binary data, with .apply(bin);

next we chunk off the first two characters with .str[2:];

then we obtain the last three characters with .str[-3:]; and

finally we again interpret these as ints, with .apply(int, base=2).

So:

>>> df.Data.apply(int, base=16).apply(bin).str[2:].str[-3:].apply(int, base=2)

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

We can however use another strategy here:

we first convert the hexadecimal number to an int; and

then we apply a bitwise and with 0b111.

for example:

>>> df.Data.apply(int, base=16) & 0b111

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

The second attempt is not only simpler, but faster as well, approximately by 66%:

>>> timeit(first_strategy, number=10000)

6.962630775000434

>>> timeit(second_strategy, number=10000)

2.330652763019316

for a dataframe that repeats the sample data 100 times, we get:

>>> timeit(first_strategy, number=10000)

17.603060900000855

>>> timeit(second_strategy, number=10000)

5.901462858979357

this is again 66% faster.

edited Nov 11 at 21:35

answered Nov 11 at 21:16

Willem Van Onsem

143k16135227

1

I like the other strategy... I might be tempted to use 0b111 instead of 7 (or possibly (1 << N) - 1 if it's going to be "the last N bits" where N is not just a couple ) or something but...
– Jon Clements♦
Nov 11 at 21:18

@JonClements: thanks for the suggestion, edited.
– Willem Van Onsem
Nov 11 at 21:26

Thank you very much. It works...but if I have to extract 3 bits from the 2nd to 5th position in a byte (for example, from "10010110" it is "010"), what I have to do? For the first solution, I could put str[p1:p2] and use None if it is the end of a string. But for the second solution, which by the way I like very much, how to extract bits that I want?
– slobokv83
Nov 12 at 7:50

@slobokv83: you can first divide it by (1<<2) (so 4), like (df.Data.apply(int, base=16) // (1 << 2)) & 0b111
– Willem Van Onsem
Nov 12 at 11:10

1

@slobokv83: well as said, you can use (df.Data.apply(int, base=16) // (1 << o)) & ((1 << w) - 1) to obtain the values between the bits o and o+w, no need to slice, zfill`, etc.
– Willem Van Onsem
Nov 12 at 12:05

|
show 2 more comments

You can use:

df.Data.apply(lambda v: int(format(int(v, 16), '08b')[-3:], 2))

Which gives you:

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

Those steps are:

Take your original data and convert it to decimal using int(number, 16) (base 16 is hex) (int('1A', 16) == 26)

Take that number and format it as a binary string format(number, '08b') gives you an character string of 0/1's zero filled on the left (format(26, '08b') == '00011010')

Take the last 3 characters of that string [-3:] ('010') and convert it to decimal with a base 2, int(binary_string[-3:], 2) gives you: 2

answered Nov 11 at 21:12

Jon Clements♦

98.2k19173218

Thank you very much. It works fine...I have another question - if I need for the length of the bits array to be variable, what I have to do? You put '08b', and if I want in some moment to be '16b', should I put like '{length}b' and for length give some value?
– slobokv83
Nov 12 at 7:58

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

We can do this by a chain of actions:

first we convert the hexadecimal number to an int with .apply(int, base=16);

next we convert this to binary data, with .apply(bin);

next we chunk off the first two characters with .str[2:];

then we obtain the last three characters with .str[-3:]; and

finally we again interpret these as ints, with .apply(int, base=2).

So:

>>> df.Data.apply(int, base=16).apply(bin).str[2:].str[-3:].apply(int, base=2)

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

We can however use another strategy here:

we first convert the hexadecimal number to an int; and

then we apply a bitwise and with 0b111.

for example:

>>> df.Data.apply(int, base=16) & 0b111

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

The second attempt is not only simpler, but faster as well, approximately by 66%:

>>> timeit(first_strategy, number=10000)

6.962630775000434

>>> timeit(second_strategy, number=10000)

2.330652763019316

for a dataframe that repeats the sample data 100 times, we get:

>>> timeit(first_strategy, number=10000)

17.603060900000855

>>> timeit(second_strategy, number=10000)

5.901462858979357

this is again 66% faster.

edited Nov 11 at 21:35

answered Nov 11 at 21:16

Willem Van Onsem

143k16135227

1

I like the other strategy... I might be tempted to use 0b111 instead of 7 (or possibly (1 << N) - 1 if it's going to be "the last N bits" where N is not just a couple ) or something but...
– Jon Clements♦
Nov 11 at 21:18

@JonClements: thanks for the suggestion, edited.
– Willem Van Onsem
Nov 11 at 21:26

Thank you very much. It works...but if I have to extract 3 bits from the 2nd to 5th position in a byte (for example, from "10010110" it is "010"), what I have to do? For the first solution, I could put str[p1:p2] and use None if it is the end of a string. But for the second solution, which by the way I like very much, how to extract bits that I want?
– slobokv83
Nov 12 at 7:50

@slobokv83: you can first divide it by (1<<2) (so 4), like (df.Data.apply(int, base=16) // (1 << 2)) & 0b111
– Willem Van Onsem
Nov 12 at 11:10

1

@slobokv83: well as said, you can use (df.Data.apply(int, base=16) // (1 << o)) & ((1 << w) - 1) to obtain the values between the bits o and o+w, no need to slice, zfill`, etc.
– Willem Van Onsem
Nov 12 at 12:05

|
show 2 more comments

We can do this by a chain of actions:

first we convert the hexadecimal number to an int with .apply(int, base=16);

next we convert this to binary data, with .apply(bin);

next we chunk off the first two characters with .str[2:];

then we obtain the last three characters with .str[-3:]; and

finally we again interpret these as ints, with .apply(int, base=2).

So:

>>> df.Data.apply(int, base=16).apply(bin).str[2:].str[-3:].apply(int, base=2)

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

We can however use another strategy here:

we first convert the hexadecimal number to an int; and

then we apply a bitwise and with 0b111.

for example:

>>> df.Data.apply(int, base=16) & 0b111

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

The second attempt is not only simpler, but faster as well, approximately by 66%:

>>> timeit(first_strategy, number=10000)

6.962630775000434

>>> timeit(second_strategy, number=10000)

2.330652763019316

for a dataframe that repeats the sample data 100 times, we get:

>>> timeit(first_strategy, number=10000)

17.603060900000855

>>> timeit(second_strategy, number=10000)

5.901462858979357

this is again 66% faster.

edited Nov 11 at 21:35

answered Nov 11 at 21:16

Willem Van Onsem

143k16135227

1

I like the other strategy... I might be tempted to use 0b111 instead of 7 (or possibly (1 << N) - 1 if it's going to be "the last N bits" where N is not just a couple ) or something but...
– Jon Clements♦
Nov 11 at 21:18

@JonClements: thanks for the suggestion, edited.
– Willem Van Onsem
Nov 11 at 21:26

Thank you very much. It works...but if I have to extract 3 bits from the 2nd to 5th position in a byte (for example, from "10010110" it is "010"), what I have to do? For the first solution, I could put str[p1:p2] and use None if it is the end of a string. But for the second solution, which by the way I like very much, how to extract bits that I want?
– slobokv83
Nov 12 at 7:50

@slobokv83: you can first divide it by (1<<2) (so 4), like (df.Data.apply(int, base=16) // (1 << 2)) & 0b111
– Willem Van Onsem
Nov 12 at 11:10

1

@slobokv83: well as said, you can use (df.Data.apply(int, base=16) // (1 << o)) & ((1 << w) - 1) to obtain the values between the bits o and o+w, no need to slice, zfill`, etc.
– Willem Van Onsem
Nov 12 at 12:05

|
show 2 more comments

We can do this by a chain of actions:

first we convert the hexadecimal number to an int with .apply(int, base=16);

next we convert this to binary data, with .apply(bin);

next we chunk off the first two characters with .str[2:];

then we obtain the last three characters with .str[-3:]; and

finally we again interpret these as ints, with .apply(int, base=2).

So:

>>> df.Data.apply(int, base=16).apply(bin).str[2:].str[-3:].apply(int, base=2)

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

We can however use another strategy here:

we first convert the hexadecimal number to an int; and

then we apply a bitwise and with 0b111.

for example:

>>> df.Data.apply(int, base=16) & 0b111

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

The second attempt is not only simpler, but faster as well, approximately by 66%:

>>> timeit(first_strategy, number=10000)

6.962630775000434

>>> timeit(second_strategy, number=10000)

2.330652763019316

for a dataframe that repeats the sample data 100 times, we get:

>>> timeit(first_strategy, number=10000)

17.603060900000855

>>> timeit(second_strategy, number=10000)

5.901462858979357

this is again 66% faster.

edited Nov 11 at 21:35

answered Nov 11 at 21:16

Willem Van Onsem

143k16135227

We can do this by a chain of actions:

first we convert the hexadecimal number to an int with .apply(int, base=16);

next we convert this to binary data, with .apply(bin);

next we chunk off the first two characters with .str[2:];

then we obtain the last three characters with .str[-3:]; and

finally we again interpret these as ints, with .apply(int, base=2).

So:

>>> df.Data.apply(int, base=16).apply(bin).str[2:].str[-3:].apply(int, base=2)

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

We can however use another strategy here:

we first convert the hexadecimal number to an int; and

then we apply a bitwise and with 0b111.

for example:

>>> df.Data.apply(int, base=16) & 0b111

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

The second attempt is not only simpler, but faster as well, approximately by 66%:

>>> timeit(first_strategy, number=10000)

6.962630775000434

>>> timeit(second_strategy, number=10000)

2.330652763019316

for a dataframe that repeats the sample data 100 times, we get:

>>> timeit(first_strategy, number=10000)

17.603060900000855

>>> timeit(second_strategy, number=10000)

5.901462858979357

this is again 66% faster.

edited Nov 11 at 21:35

answered Nov 11 at 21:16

Willem Van Onsem

143k16135227

edited Nov 11 at 21:35

answered Nov 11 at 21:16

Willem Van Onsem

143k16135227

answered Nov 11 at 21:16

Willem Van Onsem

143k16135227

answered Nov 11 at 21:16

Willem Van Onsem

143k16135227

1

I like the other strategy... I might be tempted to use 0b111 instead of 7 (or possibly (1 << N) - 1 if it's going to be "the last N bits" where N is not just a couple ) or something but...
– Jon Clements♦
Nov 11 at 21:18

@JonClements: thanks for the suggestion, edited.
– Willem Van Onsem
Nov 11 at 21:26

Thank you very much. It works...but if I have to extract 3 bits from the 2nd to 5th position in a byte (for example, from "10010110" it is "010"), what I have to do? For the first solution, I could put str[p1:p2] and use None if it is the end of a string. But for the second solution, which by the way I like very much, how to extract bits that I want?
– slobokv83
Nov 12 at 7:50

@slobokv83: you can first divide it by (1<<2) (so 4), like (df.Data.apply(int, base=16) // (1 << 2)) & 0b111
– Willem Van Onsem
Nov 12 at 11:10

1

@slobokv83: well as said, you can use (df.Data.apply(int, base=16) // (1 << o)) & ((1 << w) - 1) to obtain the values between the bits o and o+w, no need to slice, zfill`, etc.
– Willem Van Onsem
Nov 12 at 12:05

|
show 2 more comments

1

I like the other strategy... I might be tempted to use 0b111 instead of 7 (or possibly (1 << N) - 1 if it's going to be "the last N bits" where N is not just a couple ) or something but...
– Jon Clements♦
Nov 11 at 21:18

@JonClements: thanks for the suggestion, edited.
– Willem Van Onsem
Nov 11 at 21:26

Thank you very much. It works...but if I have to extract 3 bits from the 2nd to 5th position in a byte (for example, from "10010110" it is "010"), what I have to do? For the first solution, I could put str[p1:p2] and use None if it is the end of a string. But for the second solution, which by the way I like very much, how to extract bits that I want?
– slobokv83
Nov 12 at 7:50

@slobokv83: you can first divide it by (1<<2) (so 4), like (df.Data.apply(int, base=16) // (1 << 2)) & 0b111
– Willem Van Onsem
Nov 12 at 11:10

1

@slobokv83: well as said, you can use (df.Data.apply(int, base=16) // (1 << o)) & ((1 << w) - 1) to obtain the values between the bits o and o+w, no need to slice, zfill`, etc.
– Willem Van Onsem
Nov 12 at 12:05

I like the other strategy... I might be tempted to use 0b111 instead of 7 (or possibly (1 << N) - 1 if it's going to be "the last N bits" where N is not just a couple ) or something but...
– Jon Clements♦
Nov 11 at 21:18

@JonClements: thanks for the suggestion, edited.
– Willem Van Onsem
Nov 11 at 21:26

Thank you very much. It works...but if I have to extract 3 bits from the 2nd to 5th position in a byte (for example, from "10010110" it is "010"), what I have to do? For the first solution, I could put str[p1:p2] and use None if it is the end of a string. But for the second solution, which by the way I like very much, how to extract bits that I want?
– slobokv83
Nov 12 at 7:50

@slobokv83: you can first divide it by (1<<2) (so 4), like (df.Data.apply(int, base=16) // (1 << 2)) & 0b111
– Willem Van Onsem
Nov 12 at 11:10

@slobokv83: well as said, you can use (df.Data.apply(int, base=16) // (1 << o)) & ((1 << w) - 1) to obtain the values between the bits o and o+w, no need to slice, zfill`, etc.
– Willem Van Onsem
Nov 12 at 12:05

|
show 2 more comments

You can use:

df.Data.apply(lambda v: int(format(int(v, 16), '08b')[-3:], 2))

Which gives you:

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

Those steps are:

Take your original data and convert it to decimal using int(number, 16) (base 16 is hex) (int('1A', 16) == 26)

Take that number and format it as a binary string format(number, '08b') gives you an character string of 0/1's zero filled on the left (format(26, '08b') == '00011010')

Take the last 3 characters of that string [-3:] ('010') and convert it to decimal with a base 2, int(binary_string[-3:], 2) gives you: 2

answered Nov 11 at 21:12

Jon Clements♦

98.2k19173218

Thank you very much. It works fine...I have another question - if I need for the length of the bits array to be variable, what I have to do? You put '08b', and if I want in some moment to be '16b', should I put like '{length}b' and for length give some value?
– slobokv83
Nov 12 at 7:58

add a comment |

You can use:

df.Data.apply(lambda v: int(format(int(v, 16), '08b')[-3:], 2))

Which gives you:

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

Those steps are:

Take your original data and convert it to decimal using int(number, 16) (base 16 is hex) (int('1A', 16) == 26)

Take that number and format it as a binary string format(number, '08b') gives you an character string of 0/1's zero filled on the left (format(26, '08b') == '00011010')

Take the last 3 characters of that string [-3:] ('010') and convert it to decimal with a base 2, int(binary_string[-3:], 2) gives you: 2

answered Nov 11 at 21:12

Jon Clements♦

98.2k19173218

Thank you very much. It works fine...I have another question - if I need for the length of the bits array to be variable, what I have to do? You put '08b', and if I want in some moment to be '16b', should I put like '{length}b' and for length give some value?
– slobokv83
Nov 12 at 7:58

add a comment |

You can use:

df.Data.apply(lambda v: int(format(int(v, 16), '08b')[-3:], 2))

Which gives you:

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

Those steps are:

Take your original data and convert it to decimal using int(number, 16) (base 16 is hex) (int('1A', 16) == 26)

Take that number and format it as a binary string format(number, '08b') gives you an character string of 0/1's zero filled on the left (format(26, '08b') == '00011010')

Take the last 3 characters of that string [-3:] ('010') and convert it to decimal with a base 2, int(binary_string[-3:], 2) gives you: 2

answered Nov 11 at 21:12

Jon Clements♦

98.2k19173218

You can use:

df.Data.apply(lambda v: int(format(int(v, 16), '08b')[-3:], 2))

Which gives you:

0    2

1    3

2    3

3    7

4    7

5    0

6    3

Name: Data, dtype: int64

Those steps are:

Take your original data and convert it to decimal using int(number, 16) (base 16 is hex) (int('1A', 16) == 26)

Take that number and format it as a binary string format(number, '08b') gives you an character string of 0/1's zero filled on the left (format(26, '08b') == '00011010')

Take the last 3 characters of that string [-3:] ('010') and convert it to decimal with a base 2, int(binary_string[-3:], 2) gives you: 2

answered Nov 11 at 21:12

Jon Clements♦

98.2k19173218

answered Nov 11 at 21:12

Jon Clements♦

98.2k19173218

answered Nov 11 at 21:12

Jon Clements♦

98.2k19173218

answered Nov 11 at 21:12

Jon Clements♦

98.2k19173218

Thank you very much. It works fine...I have another question - if I need for the length of the bits array to be variable, what I have to do? You put '08b', and if I want in some moment to be '16b', should I put like '{length}b' and for length give some value?
– slobokv83
Nov 12 at 7:58

add a comment |

Thank you very much. It works fine...I have another question - if I need for the length of the bits array to be variable, what I have to do? You put '08b', and if I want in some moment to be '16b', should I put like '{length}b' and for length give some value?
– slobokv83
Nov 12 at 7:58

Thank you very much. It works fine...I have another question - if I need for the length of the bits array to be variable, what I have to do? You put '08b', and if I want in some moment to be '16b', should I put like '{length}b' and for length give some value?
– slobokv83
Nov 12 at 7:58

add a comment |

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