regex

问题 I want to substitute all space characters (except \n ) with "" . I tried using regular expression with \s+ but it matches with newline character as well. Is there any method to skip \n in \s in regex? 回答1: If you do not have to think of Unicode, you could use [ \t\r\f\v] Or, since \v matches a VT (verical symbol, \x0b ), \r is also considered a line break, and \f is also a kind of a vertical whitespace (rather obsolete now though - (form feed, \x0c ): [ \t] See docs: \s When the UNICODE flag

问题 I want to substitute all space characters (except \n ) with "" . I tried using regular expression with \s+ but it matches with newline character as well. Is there any method to skip \n in \s in regex? 回答1: If you do not have to think of Unicode, you could use [ \t\r\f\v] Or, since \v matches a VT (verical symbol, \x0b ), \r is also considered a line break, and \f is also a kind of a vertical whitespace (rather obsolete now though - (form feed, \x0c ): [ \t] See docs: \s When the UNICODE flag

问题 I have a DataFrame Column with one long string I would like to Parse. I am new to regex and have not worked with it yet. What I have below only returns the first name.. at best. I am wondering if parsing this string is easier for regex or creating a dictionary to iterate through. Here is what I have at the moment. The order is not always the same (C,W,D,G,UTIL) and I will be writing a for loop to iterate through multiple rows just like this one. import pandas as pd import numpy as np import

问题 I have a DataFrame Column with one long string I would like to Parse. I am new to regex and have not worked with it yet. What I have below only returns the first name.. at best. I am wondering if parsing this string is easier for regex or creating a dictionary to iterate through. Here is what I have at the moment. The order is not always the same (C,W,D,G,UTIL) and I will be writing a for loop to iterate through multiple rows just like this one. import pandas as pd import numpy as np import

问题 My regex: ^\+?(-?)0*([[:digit:]]+,[[:digit:]]+?)0*$ It is removing leading + and leading and tailing 0s in decimal number. I have tested it in regex101 For input: +000099,8420000 and substitution \1\2 it returns 99,842 I want the same result in Oracle database 11g: select REGEXP_REPLACE('+000099,8420000','^\+?(-?)0*([[:digit:]]+,[[:digit:]]+?)0*$','\1\2') from dual; But it returns 99,8420000 (tailing 0s are still present...) What I'm missing? EDIT It works like greedy quantifier * at the end

问题 My regex: ^\+?(-?)0*([[:digit:]]+,[[:digit:]]+?)0*$ It is removing leading + and leading and tailing 0s in decimal number. I have tested it in regex101 For input: +000099,8420000 and substitution \1\2 it returns 99,842 I want the same result in Oracle database 11g: select REGEXP_REPLACE('+000099,8420000','^\+?(-?)0*([[:digit:]]+,[[:digit:]]+?)0*$','\1\2') from dual; But it returns 99,8420000 (tailing 0s are still present...) What I'm missing? EDIT It works like greedy quantifier * at the end

问题 My regex: ^\+?(-?)0*([[:digit:]]+,[[:digit:]]+?)0*$ It is removing leading + and leading and tailing 0s in decimal number. I have tested it in regex101 For input: +000099,8420000 and substitution \1\2 it returns 99,842 I want the same result in Oracle database 11g: select REGEXP_REPLACE('+000099,8420000','^\+?(-?)0*([[:digit:]]+,[[:digit:]]+?)0*$','\1\2') from dual; But it returns 99,8420000 (tailing 0s are still present...) What I'm missing? EDIT It works like greedy quantifier * at the end

问题 My regex: ^\+?(-?)0*([[:digit:]]+,[[:digit:]]+?)0*$ It is removing leading + and leading and tailing 0s in decimal number. I have tested it in regex101 For input: +000099,8420000 and substitution \1\2 it returns 99,842 I want the same result in Oracle database 11g: select REGEXP_REPLACE('+000099,8420000','^\+?(-?)0*([[:digit:]]+,[[:digit:]]+?)0*$','\1\2') from dual; But it returns 99,8420000 (tailing 0s are still present...) What I'm missing? EDIT It works like greedy quantifier * at the end

问题 I would like to remove non-alpha characters from a string and convert each word into a list component such that: "All, the above." -> ["all", "the", "above"] It would seem that the following function works: re.split('\W+', str) but it does not account for corner cases. For example: "The U.S. is where it's nice." -> ["the", "U", "S", "is", "where", "it", "s", "nice"] I want the period removed but neither the apostrophe or the periods in "U.S." My idea is to create a regex where spaces are

问题 I would like to remove non-alpha characters from a string and convert each word into a list component such that: "All, the above." -> ["all", "the", "above"] It would seem that the following function works: re.split('\W+', str) but it does not account for corner cases. For example: "The U.S. is where it's nice." -> ["the", "U", "S", "is", "where", "it", "s", "nice"] I want the period removed but neither the apostrophe or the periods in "U.S." My idea is to create a regex where spaces are