Gsyfuy

Parsing a string of key-value pairs as a dictionary

Is it a fallacy if someone claims they need an explanation for every word of your argument to the point where they don't understand common terms?

Contest math problem about crossing out numbers in the table

Notes in a lick that don't fit in the scale associated with the chord

What to do when being responsible for data protection in your lab, yet advice is ignored?

How would a Dictatorship make a country more successful?

Groups acting on trees

Can a person refuse a presidential pardon?

How to avoid being sexist when trying to employ someone to function in a very sexist environment?

How to tag distinct options/entities without giving any an implicit priority or suggested order?

How can animals be objects of ethics without being subjects as well?

Avoiding morning and evening handshakes

Can we use the stored gravitational potential energy of a building to produce power?

Check if the digits in the number are in increasing sequence in python

What to do if authors don't respond to my serious concerns about their paper?

It took me a lot of time to make this, pls like. (YouTube Comments #1)

How to prevent cleaner from hanging my lock screen in Ubuntu 16.04

Grade 10 Analytic Geometry Question 23- Incredibly hard

If I delete my router's history can my ISP still provide it to my parents?

What kind of hardware implements Fourier transform?

What's a good word to describe a public place that looks like it wouldn't be rough?

Why does a metal block make a shrill sound but not a wooden block upon hammering?

Chess tournament winning streaks

What makes the Forgotten Realms "forgotten"?

Check if the digits in the number are in increasing sequence in python

How do I check if a list is empty?How do I check whether a file exists without exceptions?Calling an external command in PythonWhat are metaclasses in Python?How do I check if an array includes an object in JavaScript?Does Python have a ternary conditional operator?Check if a given key already exists in a dictionaryHow to get the number of elements in a list in Python?Does Python have a string 'contains' substring method?Easy interview question got harder: given numbers 1..100, find the missing number(s)

6

I was working on a problem that determines whether the digits in the numbers are in the increasing sequence. Now, the approach I took to solve the problem was, For instance, consider the number 5678.

To check whether 5678 is an increasing sequence, I took the first digit and the next digit and the last digit which is 5,6,8 and substitute in range function range(first,last,(diff of first digit and the next to first digit)) i.e range(5,8+1,abs(5-6)).The result is the list of digits in the ascending order

To this problem, there is a constraint saying

For incrementing sequences, 0 should come after 9, and not before 1, as in 7890. Now my program breaks at the input 7890. I don't know how to encode this logic. Can someone help me, please?.

The code for increasing sequence was

  len(set(['5','6','7','8']) - set(map(str,range(5,8+1,abs(5-6))))) == 0 

python algorithm

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Does each digit have to be exactly one bigger than the last?
  
  – John Gordon
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes @JohnGordon
  
  – s326280
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The accepted answer currently seems to fail for 78901.
  
  – גלעד ברקן
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, i didn't notice that !!.
  
  – s326280
  1 hour ago
  

  
  
  
  

add a comment | 

6

I was working on a problem that determines whether the digits in the numbers are in the increasing sequence. Now, the approach I took to solve the problem was, For instance, consider the number 5678.

To check whether 5678 is an increasing sequence, I took the first digit and the next digit and the last digit which is 5,6,8 and substitute in range function range(first,last,(diff of first digit and the next to first digit)) i.e range(5,8+1,abs(5-6)).The result is the list of digits in the ascending order

To this problem, there is a constraint saying

For incrementing sequences, 0 should come after 9, and not before 1, as in 7890. Now my program breaks at the input 7890. I don't know how to encode this logic. Can someone help me, please?.

The code for increasing sequence was

  len(set(['5','6','7','8']) - set(map(str,range(5,8+1,abs(5-6))))) == 0 

python algorithm

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Does each digit have to be exactly one bigger than the last?
  
  – John Gordon
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes @JohnGordon
  
  – s326280
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The accepted answer currently seems to fail for 78901.
  
  – גלעד ברקן
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, i didn't notice that !!.
  
  – s326280
  1 hour ago
  

  
  
  
  

add a comment | 

I was working on a problem that determines whether the digits in the numbers are in the increasing sequence. Now, the approach I took to solve the problem was, For instance, consider the number 5678.

To check whether 5678 is an increasing sequence, I took the first digit and the next digit and the last digit which is 5,6,8 and substitute in range function range(first,last,(diff of first digit and the next to first digit)) i.e range(5,8+1,abs(5-6)).The result is the list of digits in the ascending order

To this problem, there is a constraint saying

For incrementing sequences, 0 should come after 9, and not before 1, as in 7890. Now my program breaks at the input 7890. I don't know how to encode this logic. Can someone help me, please?.

The code for increasing sequence was

  len(set(['5','6','7','8']) - set(map(str,range(5,8+1,abs(5-6))))) == 0 

python algorithm

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

  len(set(['5','6','7','8']) - set(map(str,range(5,8+1,abs(5-6))))) == 0 

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

asked 3 hours ago

s326280s326280

776

asked 3 hours ago

s326280s326280

776

5 Answers
5

active

oldest

votes

3

you could zip the string representation of the number with a shifted self and iterate on consecutive digits together. Use all to check that numbers follow, using a modulo 10 to handle the 0 case.

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  you can avoid the double str'ing and slicing by using zip(*[iter(str(num))] * 2) instead but I imagine that's got way more overhead for such use in this case anyway... just throwing it out there...
  
  – Jon Clements♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This would incorrectly return True for 78901, as the OP says "0 should come after 9, and not before 1".
  
  – blhsing
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JonClements nice but in that case I would create a string beforehand.
  
  – Jean-François Fabre
  1 hour ago
  

  
  
  
  

add a comment | 

4

You can simply check if the number, when converted to a string, is a substring of '1234567890':

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

add a comment | 

2

I would create a cycling generator and slice that:

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

This is faster than solutions that check differences between individual digits. Of course, you can sacrifice the length to make it faster:

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

add a comment | 

0

Since you already have the zip version, here is an alternative solution:

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  well, it doesn't work. and it doesn't handle the case 7890 either
  
  – Jean-François Fabre
  3 hours ago
  

  
  
  
  

add a comment | 

0

Here's my take that just looks at the digits and exits as soon as there is a discrepancy:

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

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');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54954713%2fcheck-if-the-digits-in-the-number-are-in-increasing-sequence-in-python%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

you could zip the string representation of the number with a shifted self and iterate on consecutive digits together. Use all to check that numbers follow, using a modulo 10 to handle the 0 case.

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  you can avoid the double str'ing and slicing by using zip(*[iter(str(num))] * 2) instead but I imagine that's got way more overhead for such use in this case anyway... just throwing it out there...
  
  – Jon Clements♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This would incorrectly return True for 78901, as the OP says "0 should come after 9, and not before 1".
  
  – blhsing
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JonClements nice but in that case I would create a string beforehand.
  
  – Jean-François Fabre
  1 hour ago
  

  
  
  
  

add a comment | 

3

you could zip the string representation of the number with a shifted self and iterate on consecutive digits together. Use all to check that numbers follow, using a modulo 10 to handle the 0 case.

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  you can avoid the double str'ing and slicing by using zip(*[iter(str(num))] * 2) instead but I imagine that's got way more overhead for such use in this case anyway... just throwing it out there...
  
  – Jon Clements♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This would incorrectly return True for 78901, as the OP says "0 should come after 9, and not before 1".
  
  – blhsing
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JonClements nice but in that case I would create a string beforehand.
  
  – Jean-François Fabre
  1 hour ago
  

  
  
  
  

add a comment | 

you could zip the string representation of the number with a shifted self and iterate on consecutive digits together. Use all to check that numbers follow, using a modulo 10 to handle the 0 case.

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

4

You can simply check if the number, when converted to a string, is a substring of '1234567890':

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

add a comment | 

4

You can simply check if the number, when converted to a string, is a substring of '1234567890':

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

add a comment | 

You can simply check if the number, when converted to a string, is a substring of '1234567890':

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

answered 2 hours ago

blhsingblhsing

36.2k41639

answered 2 hours ago

blhsingblhsing

36.2k41639

2

I would create a cycling generator and slice that:

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

This is faster than solutions that check differences between individual digits. Of course, you can sacrifice the length to make it faster:

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

add a comment | 

2

I would create a cycling generator and slice that:

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

This is faster than solutions that check differences between individual digits. Of course, you can sacrifice the length to make it faster:

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

add a comment | 

I would create a cycling generator and slice that:

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

This is faster than solutions that check differences between individual digits. Of course, you can sacrifice the length to make it faster:

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

0

Since you already have the zip version, here is an alternative solution:

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  well, it doesn't work. and it doesn't handle the case 7890 either
  
  – Jean-François Fabre
  3 hours ago
  

  
  
  
  

add a comment | 

0

Since you already have the zip version, here is an alternative solution:

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  well, it doesn't work. and it doesn't handle the case 7890 either
  
  – Jean-François Fabre
  3 hours ago
  

  
  
  
  

add a comment | 

Since you already have the zip version, here is an alternative solution:

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

answered 3 hours ago

khachikkhachik

21.1k54381

answered 3 hours ago

khachikkhachik

21.1k54381

0

Here's my take that just looks at the digits and exits as soon as there is a discrepancy:

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

add a comment | 

0

Here's my take that just looks at the digits and exits as soon as there is a discrepancy:

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

add a comment | 

Here's my take that just looks at the digits and exits as soon as there is a discrepancy:

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542