Slice a string in groovy
groovy:000> sliceMe = “nnYYYYYYnnnnnnnnnn” ===> nnYYYYYYnnnnnnnnnn groovy:000> sliceMe[2..7] ===> YYYYYY Note the difference in the length being 1 less.
groovy:000> sliceMe = “nnYYYYYYnnnnnnnnnn” ===> nnYYYYYYnnnnnnnnnn groovy:000> sliceMe[2..7] ===> YYYYYY Note the difference in the length being 1 less.
You are close, need indexing with str which is apply for each value of Series: data[‘Order_Date’] = data[‘Shipment ID’].str[:8] For better performance if no NaNs values: data[‘Order_Date’] = [x[:8] for x in data[‘Shipment ID’]] print (data) Shipment ID Order_Date 0 20180504-S-20000 20180504 1 20180514-S-20537 20180514 2 20180514-S-20541 20180514 3 20180514-S-20644 20180514 4 20180514-S-20644 20180514 5 … Read more
You can use rangeOfString:options:range: and set the third argument to be beyond the range of the first occurrence. For example, you can do something like this: NSRange searchRange = NSMakeRange(0,string.length); NSRange foundRange; while (searchRange.location < string.length) { searchRange.length = string.length-searchRange.location; foundRange = [string rangeOfString:substring options:0 range:searchRange]; if (foundRange.location != NSNotFound) { // found an occurrence … Read more
Just use substring_index() twice: SELECT substring_index(substring_index(licence_key, ‘contract=”, -1), “issued=’, 1) FROM table;
To understand this, the very first thing you need to know is that what is the difference between substring and subsequence substring is a continuous part or subpart of a string whereas subsequence is the part of a string or sequence, that might be continuous or not but the order of the elements is maintained … Read more
You can use a while loop with str.find to find the nth occurrence if it exists and use that position to create the new string: def nth_repl(s, sub, repl, n): find = s.find(sub) # If find is not -1 we have found at least one match for the substring i = find != -1 # … Read more
For two reasons: The string meta data (e.g. length) is stored in the same memory block as the characters, to allow one string to use part of the character data of another string would mean that you would have to allocate two memory blocks for most strings instead of one. As most strings are not … Read more
new_str = str.slice(0..(str.index(‘blah’)))
Use strncpy e.g. strncpy(dest, src + beginIndex, endIndex – beginIndex); This assumes you’ve Validated that dest is large enough. endIndex is greater than beginIndex beginIndex is less than strlen(src) endIndex is less than strlen(src)
Why the c++’s implemented string::substr() doesn’t use the KMP algorithm (and doesn’t run in O(N + M)) and runs in O(N * M)? I assume you mean find(), rather than substr() which doesn’t need to search and should run in linear time (and only because it has to copy the result into a new string). … Read more