topic Re: Python string slicing fails to return the desired data. in Python Questions

Python string slicing fails to return the desired data.

BusinessNews — Tue, 03 Jan 2023 05:29:11 GMT

I am trying to search for the sum of occurances of a substring within a string:

string = 'ABCDCDC'
sub_string = 'CDC'
for i in range(len(string)-len(sub_string)):
print(string[i:len(substring)]

I am unsure why this is my output:

ABC
BC
C

Should'nt it be:

ABC
BCD
CDC
DCD
CDC

Re: Python string slicing fails to return the desired data.

Luke_Pinner — Tue, 03 Jan 2023 07:13:27 GMT

I think you mean

print(string[i:i+len(sub_string)])

Re: Python string slicing fails to return the desired data.

JohannesBierer — Tue, 03 Jan 2023 07:27:31 GMT

In addition, if you want to count the number of overlapping sub_strings in your string, maybe this could work:

string = 'ABCDCDC' sub_string = 'CDC' results = 0 sub_len = len(sub_string) for i in range(len(string)): if string[i:i+sub_len] == sub_string: results += 1 print (results)

source:

https://stackoverflow.com/questions/8899905/count-number-of-occurrences-of-a-substring-in-a-string#comment50169124_8900059

Re: Python string slicing fails to return the desired data.

DanPatterson — Tue, 03 Jan 2023 08:08:59 GMT

Love sliding windows questions

z = np.array(list('ABCDCDC')) # -- your base string s = np.array(list('CDC')) # -- sub string for query # -- numpy magic import numpy as np sl_window = np.lib.stride_tricks.sliding_window_view(z, 3) # -- windowing whr_eq = (sl_window == s[None, :]).all(-1) # -- find out where equal position = np.nonzero(whr_eq)[0] # -- at what location sl_window array([['A', 'B', 'C'], ['B', 'C', 'D'], ['C', 'D', 'C'], ['D', 'C', 'D'], ['C', 'D', 'C']], dtype='<U1') position array([2, 4], dtype=int64) sl_window[position] array([['C', 'D', 'C'], ['C', 'D', 'C']], dtype='<U1') len(position) 2

recorded for posterity

Re: Python string slicing fails to return the desired data.

Anonymous User — Tue, 03 Jan 2023 15:16:00 GMT

Your code is doing exactly what you are telling it to do. Replace the lengths with numbers for testing and you can see it:

for i in range(7-3): # equates to for i in 4: or in each iteration: 0, 1, 2, 3 in 4: print(string[i:3]) # -> is going to be: string([0:3]) string([1:3]) string([2:3]) ... etc

You're not moving the ending position so it is always at position 3, or 'C'. With your loop:

ABC -> [0:3]
BC -> [1:3]
C -> [2:3]

There are other solutions as Dan posted, but to show the difference of your code and how adding the end position 'slides' the slice:

pairs = {} for start_index in range(len(string)): # for 1 in 7: end_index = start_index + 3 #<- move the end index if len(string[start_index:end_index]) == 3: # <- check if the length of start index and end index (+3) is more than 3. combo = string[start_index:end_index] print(combo) # <- print it # added for counting the items pairs[combo] = pairs[combo] = 1 if not pairs.get(combo) else pairs[combo] + 1 else: print(f'end of string: {string[start_index:end_index]}') print (pairs) print(f' {sub_string} count: {pairs[sub_string]}')

Gives you the results:

ABC
BCD
CDC
DCD
CDC

end of string: DC
end of string: C

{'ABC': 1, 'BCD': 1, 'CDC': 2, 'DCD': 1}
CDC count: 2