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I'm trying to implement a Reed-Solomon encoder.
I start with a list of bytearray and then I have to convert all the elements of the list into str.
So now I have this list: ["bytearray(b'XXXXXXX')"]
But I have to retrieve the value from the list: "bytearray(b'XXXXXXX')" as a bytearray: bytearray(b'XXXXXXX')...
How can I perform this conversion?
I don't think you're doing it right...
If you want to convert all list elements to str, you'd use the bytearray.decode method:
In [10]: lst = [bytearray(b'XXXXXXX')]
In [11]: newlst = [x.decode('ascii') for x in lst]
In [12]: newlst
Out[12]: ['XXXXXXX']
And the reverse of that is
In [13]: [bytearray(s, 'ascii') for s in newlst]
Out[13]: [bytearray(b'XXXXXXX')]
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I want to append the number in A to 5 if it is more than 5.
A = [1,2,3,4,5,6,7,8,9,10]
To something like this:
A = [1,2,3,4,5,5,5,5,5,5]
You can try using map
A = [1,2,3,4,5,6,7,8,9,10]
list(map(lambda x: x if x<5 else 5, A))
You can use the min function to take the smaller of each list item and 5:
[min(i, 5) for i in A]
Demo: https://replit.com/#blhsing/InsidiousDigitalIntegrationtesting
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I have a list of acc nos:
list1 = ['1234','3456','2345','5543','1344','5679','6433','3243','0089']
Output I need is a string:
print(output): '1234','3456','2345','5543','1344','5679','6433','3243','0089'
You can join all the values with ',' and then adding a ' before and after the string like this:
"'{0}'".format("','".join(list1))
>>> list1 = ['1234','3456','2345','5543','1344','5679','6433','3243','0089']
>>> print(','.join(["'{0}'".format(s) for s in list1]))
'1234','3456','2345','5543','1344','5679','6433','3243','0089'
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Say I have
A = [[1.0,2.3,1.1],[2.2,1.3,3.2]]
and I want to cast all of those numbers into just ints to have
A = [[1,2,1],[2,1,3]]
How do we do that in python?
Try list comprehension*2:
print([[int(x) for x in i] for i in A])
Or list comprehension + map:
print([list(map(int,i)) for i in A])
Or map+map:
print(list(map(lambda x: list(map(int,x)),A)))
Simple ways all return:
[[1,2,1],[2,1,3]]
Here's an approach using a list comprehension and map:
A = [[1.0,2.3,1.1],[2.2,1.3,3.2]]
print([list(map(int, i)) for i in A])
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I have some strings below:
123123|00|992|1111
222222|2222|19|922
997997|3333|922|77
How can I format like this (sort by len):
123123|1111|00|992
222222|2222|19|922
997997|3333|77|922
Using str.split & str.join with sorted
Demo:
s = """123123|00|992|1111
222222|2222|19|922
997997|3333|922|77"""
data = ["|".join(sorted(i.split("|"), key=len, reverse=True)) for i in s.split("\n")]
print( "\n".join(data) )
Output:
123123|1111|992|00
222222|2222|922|19
997997|3333|922|77
Try this
a='''123123|00|992|1111
222222|2222|19|922
997997|3333|922|77'''
'\n'.join(['|'.join(sorted(e.split("|"),key=len,reverse=True)) for e in a.split("\n")])
Output:
123123|1111|992|00
222222|2222|922|19
997997|3333|922|77
The OPs sort order isn't based on string length, as the length of 2 comes before the length of 3 (the very last two segments)
for that purpose, a custom sort mapping is needed
ranks = {6:0, 4:1, 2:2, 3:3}
text = """123123|00|992|1111
222222|2222|19|922
997997|3333|922|77"""
result = '\n'.join(['|'.join(sorted(line.split('|'),key=lambda x: ranks[len(x)]))
for line in text.split('\n')])
print(result)
# outputs:
123123|1111|00|992
222222|2222|19|922
997997|3333|77|922
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My code is :
L=['my', 'my']
and I want to split the items of this list so that the output becomes:
['my'],['my']
each item in a new list
Something like this, use a list comprehension:
In [109]: L=['my', 'my']
In [110]: [[x] for x in L]
Out[110]: [['my'], ['my']]
or may you wanted this:
In [129]: print ",".join(str(x) for x in [[x] for x in L] )
['my'],['my']
In [130]: print ",".join("[{0}]".format(x) for x in L)
[my],[my]