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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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I have the list of sorted tuples, containing floats. But when I try to output the floats (without commas and brackets) it output separate tuples BUT with brackets and commas, which I do not need.
This is the part of the code:
data=[tuple1, tuple2, tuple3, tuple4]
a=sorted(data, key = lambda x: (x[0], x[1]))
b="\n".join(map(str,a))
print(b)
try this:
data=[tuple1, tuple2, tuple3, tuple4]
s=''
for i in range(len(a)):
for j in a[i]:
s += str(j)
print(s)
Try this:
print('\n'.join(' '.join(map(str, x)) for x in a))
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How would you turn a string that looks like this
7.11,8,9:00,9:15,14:30,15:00
into this dictionary entry
{7.11 : [8, (9:00, 9:15), (14:30, 15:00)]}?
Suppose that the number of time pairs (such as 9:00,9:15 and 14:30,15:00 is unknown and you want to have them all as tuple pairs.
First split the string at the commas, then zip cluster starting from the 3rd element and put it into a dictionary:
s = "7.11,8,9:00,9:15,14:30,15:00"
ss = s.split(',')
d = {ss[0]: [ss[1]] + list(zip(*[iter(ss[2:])]*2))}
Output:
{'7.11': ['8', ('9:00', '9:15'), ('14:30', '15:00')]}
If you need to convert it from string to appropiate data types (you'll have to adapt it according to your needs), then after getting the ss list:
time_list = [datetime.datetime.strptime(t,'%H:%M').time() for t in ss[2:]]
d = {float(ss[0]): [int(ss[1])] + list(zip(*[iter(time_list)]*2))}
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import re
dic={}
dic1={}
s="ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU ASK WHAT YOU CAN DO FOR YOUR `COUNTRY"
sentence=". ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU"
sentence0=". ASK WHAT YOU CAN DO FOR YOUR COUNTRY"
sentence2=sentence.split()
sentence1=sentence.split()
for position,char in enumerate(sentence1):
dic[(char)]=(position)
for position,char in enumerate(sentence2):
dic1[(char)]=(position)
dic.update(dic1)
del dic["."]
print(dic)
pattern = re.compile(r'\b(' + '|'.join(dic.keys()) + r')\b')
result = pattern.sub(lambda x: dic[x.group()], s)
print(result)
When I run the program I have these error TypeError: sequence item 0: expected str instance, int found.How can I solve it?
You need a string for the sub function, to fix that you can either change the dictionary's values to strings:
dic1[(char)]=str(position)
or change the lambda function to:
result = pattern.sub(lambda x: str(dic[x.group()]), s)
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How can I check the items in the list on the similarities in python?
I have input list, which was created:
a = input()
list = []
list += a
so if i'll have aabbbc the result which i'll need 2a3b1c
You can use collections.Counter to do the actual counting. Then use a generator expression with some string operations to format the result as you'd like.
>>> from collections import Counter
>>> ''.join(str(v) + k for k,v in sorted(Counter('aabbbc').items()))
'2a3b1c'
Use Counter
from collections import Counter
result = Counter('aabbbc')
print(result)
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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')]