I have a string array
Str_in = {'ab_cd_a9_b5__','ab_cd_r_','ef_g','3swe_4r_2345___','swe','eds______'};
how to delete the trailing underscores in the above string array. The length of each string is not constant and the n umber of underscores may vary. The expected output string is
Str_out = {'abcda9_b5','ab_cd_r','ef_g','3swe_4r_2345','swe','eds'};
Would any one help on this issue?
It's convenient to use regex to replace these characters. The pattern of trailing _ is (number of '_' is greater than or equals 1):
_+$
So the code is:
Str_in = {'ab_cd_a9_b5__','ab_cd_r_','ef_g','3swe_4r_2345___','swe','eds______'};
Str_out = cellfun(#(x) regexprep(x, '_+$', ''), Str_in, 'UniformOutput', false);
% or do as Shai mentioned,
Str_out = regexprep(Str_in, '_+$', '');
disp(Str_out);
Output:
'ab_cd_a9_b5' 'ab_cd_r' 'ef_g' '3swe_4r_2345' 'swe' 'eds'
Related
I want to know how can I replace a given string with stars except first character of string in kotlin
For e.g i have string "Rizwan" , I want it to be R*****
You can do it with padEnd():
val name = "Rizwan"
val newName = name[0].toString().padEnd(name.length, '*')
Result:
"R*****"
Try replace center of String like Phone number. with ★:
phone.replaceRange(2 , phone.length-3 , "★".repeat(phone.length-5))
pay attention to 2 + 3 = 5 :D
Result:
"09★★★★★229"
Try replacing (?<=.). with *:
val input = "Rizwan"
val output = input.replace(Regex("(?<=.)."), "*")
println(output)
This prints:
R*****
The lookbehind (?<=.) in the regex pattern ensures that we only replace a character if at least one character precedes it. This spares the first character from being replaced.
I am not an expert in Kotlin so this may not be the best way to do it but it will work for sure.
var s = "Rizwan"
var l = s.length
val first = s[0]
s=""
while(l>1) {
s=s+"*"
l--
}
s=first+s
print(s)
Basic Algorithm..... using no library or functions
val name = "Rizwan"
val newName = name[0].toString().padEnd(name.length, '*')
Result:
"R*****"
How to create a whole list of string from one string where each string in the list containing exactly one character replacement? The string itself is consisted of only four characters (say: A, B, C, and D), so that the whole list of a string of length n would contain 3n+1 strings with exactly one character replacement.
Example:
inputstr = 'ABCD'
output = ['ABCD', 'BBCD', 'CBCD', 'DBCD', 'AACD', 'ACCD', 'ADCD', 'ABAD', 'ABBD', 'ABDD', 'ABCA', 'ABCB', 'ABCC']
I write the following python code:
strin = 'ABCD'
strout = set()
tempstr1 = ''
tempstr2 = ''
tempstr3 = ''
tempstr4 = ''
for base in range(len(strin)):
if strin[base] == 'A': #this block will be repeated for char B, C and D
tempstr1 = strin.replace(strin[base], 'A')
strout.add(tempstr1)
tempstr1 = ''
tempstr2 = strin.replace(strin[base], 'B')
strout.add(tempstr2)
tempstr2 = ''
tempstr3 = strin.replace(strin[base], 'C')
strout.add(tempseq3)
tempstr3 = ''
tempstr4 = strin.replace(strin[base], 'D')
strout.add(tempseq4)
tempstr4 = ''
return strout
and it works well as long as there is no repeated character (such as 'ABCD'). However, when the input string contains repeated character (such as 'AACD'), it will return less than 3n+1 string. I tried with 'AACD' string and it returns only 10 instead of 13 strings.
Anyone can help?
change
strout = set() ===> strout = list()
I found it. I used a slicing method to create a list of total combination of strings with one replacement.
for i in range(len(seq)):
seqxlist.append(seq[:i] + 'x' + seq[i+1:])
and after that I filter out all the x-replaced strings which are longer than the original string length:
seqxlist = [x for x in seqxlist if (len(x) == len(seq))]
Then, I changed x into any of the substitution characters:
for m in seqxlist:
tempseq1 = m.replace('x', 'A')
outseq.append(tempseq1)
tempseq2 = m.replace('x', 'B')
outseq.append(tempseq2)
tempseq3 = m.replace('x', 'C')
outseq.append(tempseq3)
tempseq4 = m.replace('x', 'D')
outseq.append(tempseq4)
This will create all the possible combinations of string replacement, but still contains duplicates. To remove duplicates, I use set() to the outseq list.
I got a string array of the format
sLine =
{
[1,1] = 13-Jul-16,10.46,100.63,15.7,54.4,55656465
[1,2] = 12-Jul-16,10.47,100.64,15.7,54.4,55656465
[1,3] = 11-Jul-16,10.48,100.65,15.7,54.4,55656465
[1,4] = 10-Jul-16,10.49,100.66,15.7,54.4,55656465
}
In which each element is a string ("13-Jul-16,10.46,100.63,15.7,54.4,55656465" is a string).
I need to convert this to 6 vectors, something like
[a b c d e f] = ...
such a way, for example, for the 1st column, it would be
a = [13-Jul-16;12-Jul-16;11-Jul-16;10-Jul-16]
I tried to use cell2mat function, but for some reason it does not separate the fields into matrix elements, but it concatenates the whole string into something like
cell2mat(sLine)
ans =
13-Jul-16,10.46,100.63,15.7,54.4,5565646512-Jul-16,10.47,100.64,15.7,54.4,5565646511-Jul-16,10.48,100.65,15.7,54.4,5565646510-Jul-16,10.49,100.66,15.7,54.4,55656465
So, how can I solve this?
Update
I got the sLine matrix following the steps
pFile = urlread('http://www.google.com/finance/historical?q=BVMF:PETR4&num=365&output=csv');
sLine = strsplit(pFile,'\n');
sLine(:,1)=[];
Update
Thanks to #Suever I could get now the column dates. So the updated last version of the code is
pFile = urlread('http://www.google.com/finance/historical?q=BVMF:PETR4&num=365&output=csv');
pFile=strtrim(pFile);
sLine = strsplit(pFile,'\n');
sLine(:,1)=[];
split_values = regexp(sLine, ',', 'split');
values = cat(1, split_values{:});
values(:,1)
Your data is all strings, therefore you will need to do some string manipulation rather than using cell2mat.
You will want to split each element at the ,characters and then concatenate the result together.
sLine = {'13-Jul-16,10.46,100.63,15.7,54.4,55656465',
'12-Jul-16,10.47,100.64,15.7,54.4,55656465',
'11-Jul-16,10.48,100.65,15.7,54.4,55656465',
'10-Jul-16,10.49,100.66,15.7,54.4,55656465'};
split_values = cellfun(#(x)strsplit(x, ','), sLine, 'uniformoutput', 0);
values = cat(1, split_values{:});
values(:,1)
% {
% [1,1] = 13-Jul-16
% [2,1] = 12-Jul-16
% [3,1] = 11-Jul-16
% [4,1] = 10-Jul-16
% }
If you want it to be more concise, we can just use regexp to split it up instead of strsplit since it can accept a cell array as input.
split_values = regexp(sLine, ',', 'split');
values = cat(1, split_values{:});
Update
The issue with the code that you've posted is that there is a trailing newline in the input and when you split on newlines the last element of your sLine cell array is empty causing your issues. You'll want to use strtrim on pFile before creating the cell array to remove trailing newlines.
sLine = strsplit(strtrim(pFile), '\n');
sLine(:,1) = [];
If my string is this: 'this is a string', how can I produce all possible combinations by joining each word with its neighboring word?
What this output would look like:
this is a string
thisis a string
thisisa string
thisisastring
thisis astring
this isa string
this isastring
this is astring
What I have tried:
s = 'this is a string'.split()
for i, l in enumerate(s):
''.join(s[0:i])+' '.join(s[i:])
This produces:
'this is a string'
'thisis a string'
'thisisa string'
'thisisastring'
I realize I need to change the s[0:i] part because it's statically anchored at 0 but I don't know how to move to the next word is while still including this in the output.
A simpler (and 3x faster than the accepted answer) way to use itertools product:
s = 'this is a string'
s2 = s.replace('%', '%%').replace(' ', '%s')
for i in itertools.product((' ', ''), repeat=s.count(' ')):
print(s2 % i)
You can also use itertools.product():
import itertools
s = 'this is a string'
words = s.split()
for t in itertools.product(range(len('01')), repeat=len(words)-1):
print(''.join([words[i]+t[i]*' ' for i in range(len(t))])+words[-1])
Well, it took me a little longer than I expected... this is actually tricker than I thought :)
The main idea:
The number of spaces when you split the string is the length or the split array - 1. In our example there are 3 spaces:
'this is a string'
^ ^ ^
We'll take a binary representation of all the options to have/not have either one of the spaces, so in our case it'll be:
000
001
011
100
101
...
and for each option we'll generate the sentence respectively, where 111 represents all 3 spaces: 'this is a string' and 000 represents no-space at all: 'thisisastring'
def binaries(n):
res = []
for x in range(n ** 2 - 1):
tmp = bin(x)
res.append(tmp.replace('0b', '').zfill(n))
return res
def generate(arr, bins):
res = []
for bin in bins:
tmp = arr[0]
i = 1
for digit in list(bin):
if digit == '1':
tmp = tmp + " " + arr[i]
else:
tmp = tmp + arr[i]
i += 1
res.append(tmp)
return res
def combinations(string):
s = string.split(' ')
bins = binaries(len(s) - 1)
res = generate(s, bins)
return res
print combinations('this is a string')
# ['thisisastring', 'thisisa string', 'thisis astring', 'thisis a string', 'this isastring', 'this isa string', 'this is astring', 'this is a string']
UPDATE:
I now see that Amadan thought of the same idea - kudos for being quicker than me to think about! Great minds think alike ;)
The easiest is to do it recursively.
Terminating condition: Schrödinger join of a single element list is that word.
Recurring condition: say that L is the Schrödinger join of all the words but the first. Then the Schrödinger join of the list consists of all elements from L with the first word directly prepended, and all elements from L with the first word prepended with an intervening space.
(Assuming you are missing thisis astring by accident. If it is deliberately, I am sure I have no idea what the question is :P )
Another, non-recursive way you can do it is to enumerate all numbers from 0 to 2^(number of words - 1) - 1, then use the binary representation of each number as a selector whether or not a space needs to be present. So, for example, the abovementioned thisis astring corresponds to 0b010, for "nospace, space, nospace".
I have a string S in Matlab. How can I replace a substring in S with some pattern P. I only know the first and the last index of substring in S. What is the approach?
How about that?
str = 'My dog is called Jim'; %// original string
a = 4; %// starting index
b = 6; %// last index
replace = 'hamster'; %// new pattern
newstr = [str(1:a-1) replace str(b+1:end)]
returns:
newstr = My hamster is called Jim
In case the pattern you want to substitute has the same number of characters as the new one, you can use simple indexing:
str(a:b) = 'cat'
returns:
str = My cat is called Jim