How to access/assert tuple key values in a python dictionary

How can I return "True" or "False" if 2 consecutive strings are in a dictionary key that is a tuple/triple?

 d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

I need an expression like:

return 'a', 'b' in d.keys()

asked Nov 25 '18 at 18:02

user3383621

118112

It is supposed to check if 2 consecutive strings are in the key, that is - string 'a' followed by string 'b'

– user3383621
Nov 25 '18 at 18:23

add a comment |

How can I return "True" or "False" if 2 consecutive strings are in a dictionary key that is a tuple/triple?

 d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

I need an expression like:

return 'a', 'b' in d.keys()

asked Nov 25 '18 at 18:02

user3383621

118112

It is supposed to check if 2 consecutive strings are in the key, that is - string 'a' followed by string 'b'

– user3383621
Nov 25 '18 at 18:23

add a comment |

How can I return "True" or "False" if 2 consecutive strings are in a dictionary key that is a tuple/triple?

 d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

I need an expression like:

return 'a', 'b' in d.keys()

asked Nov 25 '18 at 18:02

user3383621

118112

How can I return "True" or "False" if 2 consecutive strings are in a dictionary key that is a tuple/triple?

 d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

I need an expression like:

return 'a', 'b' in d.keys()

python dictionary tuples

asked Nov 25 '18 at 18:02

user3383621

118112

asked Nov 25 '18 at 18:02

user3383621

118112

asked Nov 25 '18 at 18:02

user3383621

118112

asked Nov 25 '18 at 18:02

user3383621

118112

asked Nov 25 '18 at 18:02

user3383621

118112

It is supposed to check if 2 consecutive strings are in the key, that is - string 'a' followed by string 'b'

– user3383621
Nov 25 '18 at 18:23

add a comment |

It is supposed to check if 2 consecutive strings are in the key, that is - string 'a' followed by string 'b'

– user3383621
Nov 25 '18 at 18:23

It is supposed to check if 2 consecutive strings are in the key, that is - string 'a' followed by string 'b'

– user3383621
Nov 25 '18 at 18:23

add a comment |

3 Answers
3

active

oldest

votes

You can do it with nested for loops:

def myFunc(myDict):

    myKeys = list(myDict.keys())

    for myList in myKeys:

        myPreviousElement = None

        for myElement in myList:

            if myElement == myPreviousElement:

                return True

            myPreviousElement = myElement

    return False



d = {(1, 'a', 'a') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # True



d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # False

Then you can customize return values how you prefer

answered Nov 25 '18 at 18:22

Lorenzo Fiamingo

849

I think there is a problem with this... if the triple is (‘a’, ‘b’, ‘a’) it give like they are consecutive

– Lorenzo Fiamingo
Nov 27 '18 at 9:16

add a comment |

You could pair elements for each key in the dictionary and then check if any of those pairs equals your desired result, eg:

d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}



# Check for existence of any key matching criteria

any(pair == ('a', 'b') for key in d for pair in zip(key, key[1:]))

# True



# Filter out keys/values matching criteria

{k: v for k, v in d.items() if any(p == ('a', 'b') for p in zip(k, k[1:]))}

# {(1, 'a', 'b'): 2}

answered Nov 25 '18 at 18:28

Jon Clements♦

100k19175219

add a comment |

this seems to work fine

for key in d:

        return key[1] == string_1 and key[2] == string_2

answered Nov 25 '18 at 19:05

user3383621

118112

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You can do it with nested for loops:

def myFunc(myDict):

    myKeys = list(myDict.keys())

    for myList in myKeys:

        myPreviousElement = None

        for myElement in myList:

            if myElement == myPreviousElement:

                return True

            myPreviousElement = myElement

    return False



d = {(1, 'a', 'a') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # True



d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # False

Then you can customize return values how you prefer

answered Nov 25 '18 at 18:22

Lorenzo Fiamingo

849

I think there is a problem with this... if the triple is (‘a’, ‘b’, ‘a’) it give like they are consecutive

– Lorenzo Fiamingo
Nov 27 '18 at 9:16

add a comment |

You can do it with nested for loops:

def myFunc(myDict):

    myKeys = list(myDict.keys())

    for myList in myKeys:

        myPreviousElement = None

        for myElement in myList:

            if myElement == myPreviousElement:

                return True

            myPreviousElement = myElement

    return False



d = {(1, 'a', 'a') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # True



d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # False

Then you can customize return values how you prefer

answered Nov 25 '18 at 18:22

Lorenzo Fiamingo

849

I think there is a problem with this... if the triple is (‘a’, ‘b’, ‘a’) it give like they are consecutive

– Lorenzo Fiamingo
Nov 27 '18 at 9:16

add a comment |

You can do it with nested for loops:

def myFunc(myDict):

    myKeys = list(myDict.keys())

    for myList in myKeys:

        myPreviousElement = None

        for myElement in myList:

            if myElement == myPreviousElement:

                return True

            myPreviousElement = myElement

    return False



d = {(1, 'a', 'a') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # True



d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # False

Then you can customize return values how you prefer

answered Nov 25 '18 at 18:22

Lorenzo Fiamingo

849

You can do it with nested for loops:

def myFunc(myDict):

    myKeys = list(myDict.keys())

    for myList in myKeys:

        myPreviousElement = None

        for myElement in myList:

            if myElement == myPreviousElement:

                return True

            myPreviousElement = myElement

    return False



d = {(1, 'a', 'a') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # True



d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}

print(myFunc(d)) # False

Then you can customize return values how you prefer

answered Nov 25 '18 at 18:22

Lorenzo Fiamingo

849

answered Nov 25 '18 at 18:22

Lorenzo Fiamingo

849

answered Nov 25 '18 at 18:22

Lorenzo Fiamingo

849

answered Nov 25 '18 at 18:22

Lorenzo Fiamingo

849

I think there is a problem with this... if the triple is (‘a’, ‘b’, ‘a’) it give like they are consecutive

– Lorenzo Fiamingo
Nov 27 '18 at 9:16

add a comment |

I think there is a problem with this... if the triple is (‘a’, ‘b’, ‘a’) it give like they are consecutive

– Lorenzo Fiamingo
Nov 27 '18 at 9:16

I think there is a problem with this... if the triple is (‘a’, ‘b’, ‘a’) it give like they are consecutive

– Lorenzo Fiamingo
Nov 27 '18 at 9:16

add a comment |

You could pair elements for each key in the dictionary and then check if any of those pairs equals your desired result, eg:

d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}



# Check for existence of any key matching criteria

any(pair == ('a', 'b') for key in d for pair in zip(key, key[1:]))

# True



# Filter out keys/values matching criteria

{k: v for k, v in d.items() if any(p == ('a', 'b') for p in zip(k, k[1:]))}

# {(1, 'a', 'b'): 2}

answered Nov 25 '18 at 18:28

Jon Clements♦

100k19175219

add a comment |

You could pair elements for each key in the dictionary and then check if any of those pairs equals your desired result, eg:

d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}



# Check for existence of any key matching criteria

any(pair == ('a', 'b') for key in d for pair in zip(key, key[1:]))

# True



# Filter out keys/values matching criteria

{k: v for k, v in d.items() if any(p == ('a', 'b') for p in zip(k, k[1:]))}

# {(1, 'a', 'b'): 2}

answered Nov 25 '18 at 18:28

Jon Clements♦

100k19175219

add a comment |

You could pair elements for each key in the dictionary and then check if any of those pairs equals your desired result, eg:

d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}



# Check for existence of any key matching criteria

any(pair == ('a', 'b') for key in d for pair in zip(key, key[1:]))

# True



# Filter out keys/values matching criteria

{k: v for k, v in d.items() if any(p == ('a', 'b') for p in zip(k, k[1:]))}

# {(1, 'a', 'b'): 2}

answered Nov 25 '18 at 18:28

Jon Clements♦

100k19175219

You could pair elements for each key in the dictionary and then check if any of those pairs equals your desired result, eg:

d = {(1, 'a', 'b') : 2, (4, 'c', 'd'):5}



# Check for existence of any key matching criteria

any(pair == ('a', 'b') for key in d for pair in zip(key, key[1:]))

# True



# Filter out keys/values matching criteria

{k: v for k, v in d.items() if any(p == ('a', 'b') for p in zip(k, k[1:]))}

# {(1, 'a', 'b'): 2}

answered Nov 25 '18 at 18:28

Jon Clements♦

100k19175219

answered Nov 25 '18 at 18:28

Jon Clements♦

100k19175219

answered Nov 25 '18 at 18:28

Jon Clements♦

100k19175219

answered Nov 25 '18 at 18:28

Jon Clements♦

100k19175219

add a comment |

this seems to work fine

for key in d:

        return key[1] == string_1 and key[2] == string_2

answered Nov 25 '18 at 19:05

user3383621

118112

add a comment |

this seems to work fine

for key in d:

        return key[1] == string_1 and key[2] == string_2

answered Nov 25 '18 at 19:05

user3383621

118112

add a comment |

this seems to work fine

for key in d:

        return key[1] == string_1 and key[2] == string_2

answered Nov 25 '18 at 19:05

user3383621

118112

this seems to work fine

for key in d:

        return key[1] == string_1 and key[2] == string_2

answered Nov 25 '18 at 19:05

user3383621

118112

answered Nov 25 '18 at 19:05

user3383621

118112

answered Nov 25 '18 at 19:05

user3383621

118112

answered Nov 25 '18 at 19:05

user3383621

118112

add a comment |

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