re.sub how to do it in one step?

I try to put this steps in one, but it doesnt work

w = re.sub('[0-9]', r'9', w)

w = re.sub('[A-Z]', r'X', w)

w = re.sub('[a-z]', r'x', w)

Does anybody knows how to make from such strings as XXxxxx999 --> Xx9.

edited Nov 13 '18 at 14:56

Wiktor Stribiżew

312k16132207

asked Nov 13 '18 at 13:25

Nastja Kr

687

2

Настя, you may use a callback method as the replacement argument. There, you may apply custom replacement logic, just build your regex with capturing groups. like ([0-9])|([A-Z])|[a-z], and then analyze which group matched.

– Wiktor Stribiżew
Nov 13 '18 at 13:33

I understood it would be like that, but i dont understand how to make in replacement some options w = re.sub('([0-9])|([A-Z])|([a-z])', r'9', w)

– Nastja Kr
Nov 13 '18 at 13:38

1

re.sub(r'([0-9])|([A-Z])|[a-z]', repl, w) where repl is def repl(m):...., just check if m.group(1) then it is a digit, if m.group(2) then it is an uppercase letter...

– Wiktor Stribiżew
Nov 13 '18 at 13:46

add a comment |

I try to put this steps in one, but it doesnt work

w = re.sub('[0-9]', r'9', w)

w = re.sub('[A-Z]', r'X', w)

w = re.sub('[a-z]', r'x', w)

Does anybody knows how to make from such strings as XXxxxx999 --> Xx9.

edited Nov 13 '18 at 14:56

Wiktor Stribiżew

312k16132207

asked Nov 13 '18 at 13:25

Nastja Kr

687

2

Настя, you may use a callback method as the replacement argument. There, you may apply custom replacement logic, just build your regex with capturing groups. like ([0-9])|([A-Z])|[a-z], and then analyze which group matched.

– Wiktor Stribiżew
Nov 13 '18 at 13:33

I understood it would be like that, but i dont understand how to make in replacement some options w = re.sub('([0-9])|([A-Z])|([a-z])', r'9', w)

– Nastja Kr
Nov 13 '18 at 13:38

1

re.sub(r'([0-9])|([A-Z])|[a-z]', repl, w) where repl is def repl(m):...., just check if m.group(1) then it is a digit, if m.group(2) then it is an uppercase letter...

– Wiktor Stribiżew
Nov 13 '18 at 13:46

add a comment |

I try to put this steps in one, but it doesnt work

w = re.sub('[0-9]', r'9', w)

w = re.sub('[A-Z]', r'X', w)

w = re.sub('[a-z]', r'x', w)

Does anybody knows how to make from such strings as XXxxxx999 --> Xx9.

edited Nov 13 '18 at 14:56

Wiktor Stribiżew

312k16132207

asked Nov 13 '18 at 13:25

Nastja Kr

687

I try to put this steps in one, but it doesnt work

w = re.sub('[0-9]', r'9', w)

w = re.sub('[A-Z]', r'X', w)

w = re.sub('[a-z]', r'x', w)

Does anybody knows how to make from such strings as XXxxxx999 --> Xx9.

regex python-3.x

edited Nov 13 '18 at 14:56

Wiktor Stribiżew

312k16132207

asked Nov 13 '18 at 13:25

Nastja Kr

687

edited Nov 13 '18 at 14:56

Wiktor Stribiżew

312k16132207

asked Nov 13 '18 at 13:25

Nastja Kr

687

edited Nov 13 '18 at 14:56

Wiktor Stribiżew

312k16132207

edited Nov 13 '18 at 14:56

Wiktor Stribiżew

312k16132207

edited Nov 13 '18 at 14:56

Wiktor Stribiżew

312k16132207

asked Nov 13 '18 at 13:25

Nastja Kr

687

asked Nov 13 '18 at 13:25

Nastja Kr

687

asked Nov 13 '18 at 13:25

Nastja Kr

687

2

Настя, you may use a callback method as the replacement argument. There, you may apply custom replacement logic, just build your regex with capturing groups. like ([0-9])|([A-Z])|[a-z], and then analyze which group matched.

– Wiktor Stribiżew
Nov 13 '18 at 13:33

I understood it would be like that, but i dont understand how to make in replacement some options w = re.sub('([0-9])|([A-Z])|([a-z])', r'9', w)

– Nastja Kr
Nov 13 '18 at 13:38

1

re.sub(r'([0-9])|([A-Z])|[a-z]', repl, w) where repl is def repl(m):...., just check if m.group(1) then it is a digit, if m.group(2) then it is an uppercase letter...

– Wiktor Stribiżew
Nov 13 '18 at 13:46

add a comment |

2

Настя, you may use a callback method as the replacement argument. There, you may apply custom replacement logic, just build your regex with capturing groups. like ([0-9])|([A-Z])|[a-z], and then analyze which group matched.

– Wiktor Stribiżew
Nov 13 '18 at 13:33

I understood it would be like that, but i dont understand how to make in replacement some options w = re.sub('([0-9])|([A-Z])|([a-z])', r'9', w)

– Nastja Kr
Nov 13 '18 at 13:38

1

re.sub(r'([0-9])|([A-Z])|[a-z]', repl, w) where repl is def repl(m):...., just check if m.group(1) then it is a digit, if m.group(2) then it is an uppercase letter...

– Wiktor Stribiżew
Nov 13 '18 at 13:46

Настя, you may use a callback method as the replacement argument. There, you may apply custom replacement logic, just build your regex with capturing groups. like ([0-9])|([A-Z])|[a-z], and then analyze which group matched.

– Wiktor Stribiżew
Nov 13 '18 at 13:33

I understood it would be like that, but i dont understand how to make in replacement some options w = re.sub('([0-9])|([A-Z])|([a-z])', r'9', w)

– Nastja Kr
Nov 13 '18 at 13:38

re.sub(r'([0-9])|([A-Z])|[a-z]', repl, w) where repl is def repl(m):...., just check if m.group(1) then it is a digit, if m.group(2) then it is an uppercase letter...

– Wiktor Stribiżew
Nov 13 '18 at 13:46

add a comment |

1 Answer
1

active

oldest

votes

You may use a callback method as a replacement argument like this:

import re



rx = r'([0-9]+)|([A-Z]+)|[a-z]+'

w = "XXxxxx999"



def repl(m):

    if m.group(1):       # if ([0-9]) matched

        return '9'       # replace with 9

    elif m.group(2):     # if ([A-Z]) matched

        return 'X'       # replace with X

    else:                # if ([a-z]) matched

        return 'x'       # replace with x



print(re.sub(rx, repl, w)) # => Xx9

See the Python demo.

The regex matches:

([0-9]+) - Group 1: 1+ digits

| - or

([A-Z]+) - Group 2: 1+ uppercase letters

| - or

[a-z]+ - 1+ lowercase letters.

edited Nov 13 '18 at 14:57

answered Nov 13 '18 at 14:17

Wiktor Stribiżew

312k16132207

great, thanks I am new in programming so I have learned today somthing. Can you laso help me with the second question ?I found some solutions for this problem, but only if we want to eliminate all the characters, that appear in the string, and not like in my case. input = "XXXxxx999XX" output = "Xx9X"

– Nastja Kr
Nov 13 '18 at 14:40

1

@NastjaKryvoscheya Do you mean ideone.com/iPPeVF ? Use rx = r'([0-9]+)|([A-Z]+)|[a-z]+' where + quantifier is added to match 1 or more occurrences of each subpattern.

– Wiktor Stribiżew
Nov 13 '18 at 14:47

wow great, thanks.... didnt thought it is so easy

– Nastja Kr
Nov 13 '18 at 14:53

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53282036%2fre-sub-how-to-do-it-in-one-step%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

You may use a callback method as a replacement argument like this:

import re



rx = r'([0-9]+)|([A-Z]+)|[a-z]+'

w = "XXxxxx999"



def repl(m):

    if m.group(1):       # if ([0-9]) matched

        return '9'       # replace with 9

    elif m.group(2):     # if ([A-Z]) matched

        return 'X'       # replace with X

    else:                # if ([a-z]) matched

        return 'x'       # replace with x



print(re.sub(rx, repl, w)) # => Xx9

See the Python demo.

The regex matches:

([0-9]+) - Group 1: 1+ digits

| - or

([A-Z]+) - Group 2: 1+ uppercase letters

| - or

[a-z]+ - 1+ lowercase letters.

edited Nov 13 '18 at 14:57

answered Nov 13 '18 at 14:17

Wiktor Stribiżew

312k16132207

great, thanks I am new in programming so I have learned today somthing. Can you laso help me with the second question ?I found some solutions for this problem, but only if we want to eliminate all the characters, that appear in the string, and not like in my case. input = "XXXxxx999XX" output = "Xx9X"

– Nastja Kr
Nov 13 '18 at 14:40

1

@NastjaKryvoscheya Do you mean ideone.com/iPPeVF ? Use rx = r'([0-9]+)|([A-Z]+)|[a-z]+' where + quantifier is added to match 1 or more occurrences of each subpattern.

– Wiktor Stribiżew
Nov 13 '18 at 14:47

wow great, thanks.... didnt thought it is so easy

– Nastja Kr
Nov 13 '18 at 14:53

add a comment |

You may use a callback method as a replacement argument like this:

import re



rx = r'([0-9]+)|([A-Z]+)|[a-z]+'

w = "XXxxxx999"



def repl(m):

    if m.group(1):       # if ([0-9]) matched

        return '9'       # replace with 9

    elif m.group(2):     # if ([A-Z]) matched

        return 'X'       # replace with X

    else:                # if ([a-z]) matched

        return 'x'       # replace with x



print(re.sub(rx, repl, w)) # => Xx9

See the Python demo.

The regex matches:

([0-9]+) - Group 1: 1+ digits

| - or

([A-Z]+) - Group 2: 1+ uppercase letters

| - or

[a-z]+ - 1+ lowercase letters.

edited Nov 13 '18 at 14:57

answered Nov 13 '18 at 14:17

Wiktor Stribiżew

312k16132207

great, thanks I am new in programming so I have learned today somthing. Can you laso help me with the second question ?I found some solutions for this problem, but only if we want to eliminate all the characters, that appear in the string, and not like in my case. input = "XXXxxx999XX" output = "Xx9X"

– Nastja Kr
Nov 13 '18 at 14:40

1

@NastjaKryvoscheya Do you mean ideone.com/iPPeVF ? Use rx = r'([0-9]+)|([A-Z]+)|[a-z]+' where + quantifier is added to match 1 or more occurrences of each subpattern.

– Wiktor Stribiżew
Nov 13 '18 at 14:47

wow great, thanks.... didnt thought it is so easy

– Nastja Kr
Nov 13 '18 at 14:53

add a comment |

You may use a callback method as a replacement argument like this:

import re



rx = r'([0-9]+)|([A-Z]+)|[a-z]+'

w = "XXxxxx999"



def repl(m):

    if m.group(1):       # if ([0-9]) matched

        return '9'       # replace with 9

    elif m.group(2):     # if ([A-Z]) matched

        return 'X'       # replace with X

    else:                # if ([a-z]) matched

        return 'x'       # replace with x



print(re.sub(rx, repl, w)) # => Xx9

See the Python demo.

The regex matches:

([0-9]+) - Group 1: 1+ digits

| - or

([A-Z]+) - Group 2: 1+ uppercase letters

| - or

[a-z]+ - 1+ lowercase letters.

edited Nov 13 '18 at 14:57

answered Nov 13 '18 at 14:17

Wiktor Stribiżew

312k16132207

You may use a callback method as a replacement argument like this:

import re



rx = r'([0-9]+)|([A-Z]+)|[a-z]+'

w = "XXxxxx999"



def repl(m):

    if m.group(1):       # if ([0-9]) matched

        return '9'       # replace with 9

    elif m.group(2):     # if ([A-Z]) matched

        return 'X'       # replace with X

    else:                # if ([a-z]) matched

        return 'x'       # replace with x



print(re.sub(rx, repl, w)) # => Xx9

See the Python demo.

The regex matches:

([0-9]+) - Group 1: 1+ digits

| - or

([A-Z]+) - Group 2: 1+ uppercase letters

| - or

[a-z]+ - 1+ lowercase letters.

edited Nov 13 '18 at 14:57

answered Nov 13 '18 at 14:17

Wiktor Stribiżew

312k16132207

edited Nov 13 '18 at 14:57

answered Nov 13 '18 at 14:17

Wiktor Stribiżew

312k16132207

answered Nov 13 '18 at 14:17

Wiktor Stribiżew

312k16132207

answered Nov 13 '18 at 14:17

Wiktor Stribiżew

312k16132207

great, thanks I am new in programming so I have learned today somthing. Can you laso help me with the second question ?I found some solutions for this problem, but only if we want to eliminate all the characters, that appear in the string, and not like in my case. input = "XXXxxx999XX" output = "Xx9X"

– Nastja Kr
Nov 13 '18 at 14:40

1

@NastjaKryvoscheya Do you mean ideone.com/iPPeVF ? Use rx = r'([0-9]+)|([A-Z]+)|[a-z]+' where + quantifier is added to match 1 or more occurrences of each subpattern.

– Wiktor Stribiżew
Nov 13 '18 at 14:47

wow great, thanks.... didnt thought it is so easy

– Nastja Kr
Nov 13 '18 at 14:53

add a comment |

great, thanks I am new in programming so I have learned today somthing. Can you laso help me with the second question ?I found some solutions for this problem, but only if we want to eliminate all the characters, that appear in the string, and not like in my case. input = "XXXxxx999XX" output = "Xx9X"

– Nastja Kr
Nov 13 '18 at 14:40

1

@NastjaKryvoscheya Do you mean ideone.com/iPPeVF ? Use rx = r'([0-9]+)|([A-Z]+)|[a-z]+' where + quantifier is added to match 1 or more occurrences of each subpattern.

– Wiktor Stribiżew
Nov 13 '18 at 14:47

wow great, thanks.... didnt thought it is so easy

– Nastja Kr
Nov 13 '18 at 14:53

great, thanks I am new in programming so I have learned today somthing. Can you laso help me with the second question ?I found some solutions for this problem, but only if we want to eliminate all the characters, that appear in the string, and not like in my case. input = "XXXxxx999XX" output = "Xx9X"

– Nastja Kr
Nov 13 '18 at 14:40

@NastjaKryvoscheya Do you mean ideone.com/iPPeVF ? Use rx = r'([0-9]+)|([A-Z]+)|[a-z]+' where + quantifier is added to match 1 or more occurrences of each subpattern.

– Wiktor Stribiżew
Nov 13 '18 at 14:47

wow great, thanks.... didnt thought it is so easy

– Nastja Kr
Nov 13 '18 at 14:53

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Ndtyjky