Input is a number, e.g. 9 and I want to print decimal, octal, hex and binary value from 1 to 9 like:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
How can I achieve this in python3 using syntax like
dm, oc, hx, bn = len(str(9)), len(bin(9)[2:]), ...
print("{:dm%d} {:oc%s}" % (i, oct(i[2:]))
I mean if number is 999 so I want decimal 10 to be printed like ' 10' and binary equivalent of 999 is 1111100111 so I want 10 like ' 1010'.
You can use str.format() and its mini-language to do the whole thing for you:
for i in range(1, 10):
print("{v} {v:>6o} {v:>6x} {v:>6b}".format(v=i))
Which will print:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
UPDATE: To define field 'widths' in a variable you can use a format-within-format structure:
w = 5 # field width, i.e. offset to the right for all octal/hex/binary values
for i in range(1, 10):
print("{v} {v:>{w}o} {v:>{w}x} {v:>{w}b}".format(v=i, w=w))
Or define a different width variable for each field type if you want them non-uniformly spaced.
Btw. since you've tagged your question with python-3.x, if you're using Python 3.6 or newer, you can use Literal String Interpolation to simplify it even more:
w = 5 # field width, i.e. offset to the right for all octal/hex/binary values
for v in range(1, 10):
print(f"{v} {v:>{w}o} {v:>{w}x} {v:>{w}b}")
Related
I have a dataframe such as :
Name Position Value
A 1 10
A 2 11
A 3 10
A 4 8
A 5 6
A 6 12
A 7 10
A 8 9
A 9 9
A 10 9
A 11 9
A 12 9
and I woulde like for each interval of 3 position, to calculate the mean of Values.
And create a new df with start and end coordinates (of length 3 then), with the Mean_value column.
Name Start End Mean_value
A 1 3 10.33 <---- here this is (10+11+10)/3 = 10.33
A 4 6 8.7
A 7 9 9.3
A 10 13 9
Does someone have an idea using pandas please ?
Solution for get each 3 rows (if exist) per Name groups - first get counter by GroupBy.cumcount with integer division and pass it to named aggregations:
g = df.groupby('Name').cumcount() // 3
df = df.groupby(['Name',g]).agg(Start=('Position','first'),
End=('Position','last'),
Value=('Value','mean')).droplevel(1).reset_index()
print (df)
Name Start End Value
0 A 1 3 10.333333
1 A 4 6 8.666667
2 A 7 9 9.333333
3 A 10 12 9.000000
If I type !i.10 it gives the factorial of first 10 numbers.
However if I try to add a column of ordinal numbers >/.!i.10, 1+i.10, then J freezes or I get an "Out of memory" error. How do I create custom tables?
I think that what is happening is that you are creating something much bigger than you expect. Taking it in steps:
1+ i. 10 NB. list of 1 to 10
1 2 3 4 5 6 7 8 9 10
10 , 1+ i. 10 NB. 10 prepended
10 1 2 3 4 5 6 7 8 9 10
i. 10 , 1+ i. 10 NB. creates an 11 dimension array with shape 10 1 2 3 4 5 6 7 8 9 10 and largest value of 36287999
When you apply ! to that i. 10 , 1+ i. 10 you get some very large numbers. I am not sure what you are trying to do with the leading >/.
Is this what you had in mind?
(!1 + i.10),. (1+i.10) NB. using parenthesis to isolate operations
1 1
2 2
6 3
24 4
120 5
720 6
5040 7
40320 8
362880 9
3.6288e6 10
To give extended type and get rid of the 3.6288e6 we can use x:
(x:!1 + i.10),. (1+i.10)
1 1
2 2
6 3
24 4
120 5
720 6
5040 7
40320 8
362880 9
3628800 10
or tacit
(x:#! ,. ]) # (1+i.) 10
1 1
2 2
6 3
24 4
120 5
720 6
5040 7
40320 8
362880 9
3628800 10
Or a version I find a little better
([: (,.~ !) 1x + i.) 10
1 1
2 2
6 3
24 4
120 5
720 6
5040 7
40320 8
362880 9
3628800 10
I'm trying to do this program where given a number N, one has to print out the decimal, octal, hexadecimal and binary for all the numbers in range 1 to N. The trouble is that the platform requires the solution in a particular format.
Suppose the number is 17, so the output should be like :
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
10 12 A 1010
11 13 B 1011
12 14 C 1100
13 15 D 1101
14 16 E 1110
15 17 F 1111
16 20 10 10000
17 21 11 10001
For 7 it would be like :
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
If you notice, the above is required to be printed in a way that the decimal, octal and hexadecimal numbers need a minimum of 2 spaces at their left whereas the binary numbers need at least one space at their left. Now, as the length of the numbers increase the space needs to be given accordingly such that the minimum space is there even for the max length number. So, how do I print them using a variable space? So far I have tried this :
Code
def print_formatted(number):
space=len(str(bin(number))[2:])
for i in range(1,number+1):
print('{:2d}'.format(i), end='')
print('{:>3s}'.format(str(oct(i))[2:]), end='')
print('{:>3s}'.format(str(hex(i))[2:]), end='')
print('{:>'+str(space)+'s}'.format(str(bin(i))[2:]))
print_formatted(17)
Here, I just tried doing the required with just the binary numbers but it's giving me an error
print('{:>'+str(space)+'s}'.format(str(bin(i))[2:]))
ValueError: Single '}' encountered in format string
Is there any fix/alternative for this?
Your problem is operator order - the + for string concattenation is weaker then the method call in
'{:>' + str(space) + 's}'.format(str(bin(i))[2:])
. Thats why you call the .format(...) only on "s}" - not the whole string. And thats where the
ValueError: Single '}' encountered in format string
comes from.
Putting the complete formatstring into parenthesis before applying .format to it fixes that.
You also need 1 more space for binary and can skip some str() that are not needed:
def print_formatted(number):
space=len(str(bin(number))[2:])+1 # fix here
for i in range(1,number+1):
print('{:2d}'.format(i), end='')
print('{:>3s}'.format(oct(i)[2:]), end='')
print('{:>3s}'.format(hex(i)[2:]), end='')
print(('{:>'+str(space)+'s}').format(bin(i)[2:])) # fix here
print_formatted(17)
Output:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
10 12 a 1010
11 13 b 1011
12 14 c 1100
13 15 d 1101
14 16 e 1110
15 17 f 1111
16 20 10 10000
17 21 11 10001
From your given output above you might need to prepend this by 2 spaces - not sure if its a formatting error in your output above or part of the restrictions.
You could also shorten this by using f-strings (and removing superflous str() around bin, oct, hex: they all return a strings already).
Then you need to calculate the the numbers you use to your space out your input values:
def print_formatted(number):
de,bi,oc,he = len(str(number)), len(bin(number)), len(oct(number)), len(hex(number))
for i in range(1,number+1):
print(f' {i:{de}d}{oct(i)[2:]:>{oc}s}{hex(i)[2:]:>{he}s}{bin(i)[2:]:>{bi}s}')
print_formatted(26)
to accomodate other values then 17, f.e. 128:
1 1 1 1
2 2 2 10
3 3 3 11
...
8 10 8 1000
...
16 20 10 10000
...
32 40 20 100000
...
64 100 40 1000000
...
128 200 80 10000000
Using python 3 am trying for each uniqe row in the column 'Name' to get the last 5 records from the column 'Number'. How exactly can this be done in python?
My df looks like this:
Name Number
a 5
a 6
b 7
b 8
a 9
a 10
b 11
b 12
a 9
b 8
I saw same exmples(like this one Get sum of last 5 rows for each unique id ) in SQL but that is time consuming and I would like to learn how to do it in python.
My expected output df would be like this:
Name 1 2 3 4 5
a 5 6 9 10 9
b 7 8 11 12 8
I think you need something like this:
df_out = df.groupby('Name').tail(5)
df_out.set_index(['Name', df_out.groupby('Name').cumcount() +1])['Number'].unstack()
Output:
1 2 3 4 5
Name
a 5 6 9 10 9
b 7 8 11 12 8
Looks like you need pivot after a groupby.cumcount()
df1=df.groupby('Name').tail(5)
final=(df1.assign(k=df1.groupby('Name').cumcount()+1)
.pivot(index='Name', columns='k', values='Number')
.reset_index().rename_axis(None, axis=1))
print(final)
Name 1 2 3 4 5
0 a 5 6 9 10 9
1 b 7 8 11 12 8
I have 15 datafiles with unequal row sizes, but number of columns in each file is same. e.g.
ifile1.dat ifile2.dat ifile3.dat and so on ............
0 0 0 0 1 6
1 2 5 3 2 7
2 5 6 10 4 6
5 2 8 9 5 9
10 2 10 3 8 2
In each file 1st column represents the index number.
I would like to compute average of all these files for each index number in column 1. i.e.
ofile.txt
0 0 [This is computed as (0+0)/2]
1 4 [This is computed as (2+6)/2]
2 6 [This is computed as (5+7)/2]
3 [no value]
4 6 [This is computed as (6)/1]
5 4.66 [This is computed as (2+3+9)/3]
6 10
7
8 5.5
9
10 2.5
I can't think of any simple method to do it. I was thinking of a method, but seems very lengthy. Taking the average after converting all the files with same row size, .e.g.
ifile1.dat ifile2.dat ifile3.dat and so on ............
0 0 0 0 0 0
1 2 1 1 6
2 5 2 2 7
3 3 3
4 4 4 6
5 2 5 3 5 9
6 6 10 6
7 7 7
8 8 9 8 2
9 9 9
10 2 10 3 10
$ awk '{s[$1]+=$2; c[$1]++;} END{for (i in s) print i,s[i]/c[i];}' ifile*.dat
0 0
1 4
2 6
4 6
5 4.66667
6 10
8 5.5
10 2.5
In the above code, there are two arrays, s and c. s[i] is the sum of all entries with index i and c[i] is the number of entries with index i. After we have read all the files, we print the average, s[i]/c[i], for each index i.