How to slice a list containing a set of strings by comma?

I have a set of strings inside a list that I read from a CSV file and it looks like this:

myList = ('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106'),...

I want to store it into a database (Django) that has three fields:

fieldOne = '17.0.1.34', '203.84.210.248','202.73.40.45', ....

fieldTwo = '17.1.182.21', '27.111.228.3', '27.111.228.6', ...

fieldThree = '714', '10310', '18106',...

What should I do?

edited Nov 20 '18 at 3:46

Mihai Chelaru

2,180101022

asked Nov 20 '18 at 3:44

Banthita Limwilai

213

2

Are fieldOne, fieldTwo etc. lists of strings or strings separated by a comma? If they're lists, please surround them with square brackets to make it clearer.

– slider
Nov 20 '18 at 3:47

add a comment |

I have a set of strings inside a list that I read from a CSV file and it looks like this:

myList = ('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106'),...

I want to store it into a database (Django) that has three fields:

fieldOne = '17.0.1.34', '203.84.210.248','202.73.40.45', ....

fieldTwo = '17.1.182.21', '27.111.228.3', '27.111.228.6', ...

fieldThree = '714', '10310', '18106',...

What should I do?

edited Nov 20 '18 at 3:46

Mihai Chelaru

2,180101022

asked Nov 20 '18 at 3:44

Banthita Limwilai

213

2

Are fieldOne, fieldTwo etc. lists of strings or strings separated by a comma? If they're lists, please surround them with square brackets to make it clearer.

– slider
Nov 20 '18 at 3:47

add a comment |

I have a set of strings inside a list that I read from a CSV file and it looks like this:

myList = ('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106'),...

I want to store it into a database (Django) that has three fields:

fieldOne = '17.0.1.34', '203.84.210.248','202.73.40.45', ....

fieldTwo = '17.1.182.21', '27.111.228.3', '27.111.228.6', ...

fieldThree = '714', '10310', '18106',...

What should I do?

edited Nov 20 '18 at 3:46

Mihai Chelaru

2,180101022

asked Nov 20 '18 at 3:44

Banthita Limwilai

213

I have a set of strings inside a list that I read from a CSV file and it looks like this:

myList = ('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106'),...

I want to store it into a database (Django) that has three fields:

fieldOne = '17.0.1.34', '203.84.210.248','202.73.40.45', ....

fieldTwo = '17.1.182.21', '27.111.228.3', '27.111.228.6', ...

fieldThree = '714', '10310', '18106',...

What should I do?

python python-3.x

edited Nov 20 '18 at 3:46

Mihai Chelaru

2,180101022

asked Nov 20 '18 at 3:44

Banthita Limwilai

213

edited Nov 20 '18 at 3:46

Mihai Chelaru

2,180101022

asked Nov 20 '18 at 3:44

Banthita Limwilai

213

edited Nov 20 '18 at 3:46

Mihai Chelaru

2,180101022

edited Nov 20 '18 at 3:46

Mihai Chelaru

2,180101022

edited Nov 20 '18 at 3:46

Mihai Chelaru

2,180101022

asked Nov 20 '18 at 3:44

Banthita Limwilai

213

asked Nov 20 '18 at 3:44

Banthita Limwilai

213

asked Nov 20 '18 at 3:44

Banthita Limwilai

213

2

Are fieldOne, fieldTwo etc. lists of strings or strings separated by a comma? If they're lists, please surround them with square brackets to make it clearer.

– slider
Nov 20 '18 at 3:47

add a comment |

2

Are fieldOne, fieldTwo etc. lists of strings or strings separated by a comma? If they're lists, please surround them with square brackets to make it clearer.

– slider
Nov 20 '18 at 3:47

Are fieldOne, fieldTwo etc. lists of strings or strings separated by a comma? If they're lists, please surround them with square brackets to make it clearer.

– slider
Nov 20 '18 at 3:47

add a comment |

4 Answers
4

active

oldest

votes

You have a list of 3-tuples:

myList = [

    (a1, b1, c1),

    (a2, b2, c2),

    (a3, b3, c3),

    ...

]

You want to refactor this list into three variables:

fieldOne = [a1, a2, a3, ...]

fieldTwo = [b1, b2, b3, ...]

fieldThree = [c1, c2, c3, ...]

You can do this using list comprehension: "Make a list by taking the kth value from each element of myList". That would look like this:

fieldOne = [i[0] for i in myList]    # k = 0

fieldTwo = [i[1] for i in myList]    # k = 1

fieldThree = [i[2] for i in myList]  # k = 2

Of course, if your tuples are of variable length this gets much more complicated - but these are the fundamentals, and you can play around with them as needed to solve your problem.

answered Nov 20 '18 at 3:49

Green Cloak Guy

2,5131720

add a comment |

You could also zip() your results and store them in a dictionary:

>>> myList = [('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106')]

>>> fields = 'fieldOne', 'fieldTwo', 'fieldThree'

>>> dict(zip(fields, zip(*myList)))

{'fieldOne': ('17.0.1.34', '203.84.210.248', '202.73.40.45'), 'fieldTwo': ('17.1.182.21', '27.111.228.3', '27.111.228.6'), 'fieldThree': ('714', '10310', '18106')}

Then you can just reference the 'fieldOne', 'fieldTwo', 'fieldThree' keys from this dictionary and store it in your database.

answered Nov 20 '18 at 3:57

RoadRunner

11.1k31340

add a comment |

I would suggest using the following way so you can use it multiple times.

def create_field(myList):

    fieldOne = [myList[i][0] for i in range(len(myList))]

    fieldTwo = [myList[i][1] for i in range(len(myList))]

    fieldThree = [myList[i][2] for i in range(len(myList))]

    return fieldOne, fieldTwo, fieldThree

create_field(myList)

answered Nov 20 '18 at 4:05

hemanta

178

add a comment |

Given your data, here is a possible solution:

# Initialize each field list

fieldOne = 

fieldTwo = 

fieldThree = 



# fetch each **item** in **myList**, we extract the component corresponding

# append it to the appropriate field

for item in myList:

    fieldOne.append(item[0])

    fieldTwo.append(item[1])

    fieldThree.append(item[2])

edited Nov 20 '18 at 4:25

answered Nov 20 '18 at 4:00

eapetcho

42927

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

You have a list of 3-tuples:

myList = [

    (a1, b1, c1),

    (a2, b2, c2),

    (a3, b3, c3),

    ...

]

You want to refactor this list into three variables:

fieldOne = [a1, a2, a3, ...]

fieldTwo = [b1, b2, b3, ...]

fieldThree = [c1, c2, c3, ...]

You can do this using list comprehension: "Make a list by taking the kth value from each element of myList". That would look like this:

fieldOne = [i[0] for i in myList]    # k = 0

fieldTwo = [i[1] for i in myList]    # k = 1

fieldThree = [i[2] for i in myList]  # k = 2

Of course, if your tuples are of variable length this gets much more complicated - but these are the fundamentals, and you can play around with them as needed to solve your problem.

answered Nov 20 '18 at 3:49

Green Cloak Guy

2,5131720

add a comment |

You have a list of 3-tuples:

myList = [

    (a1, b1, c1),

    (a2, b2, c2),

    (a3, b3, c3),

    ...

]

You want to refactor this list into three variables:

fieldOne = [a1, a2, a3, ...]

fieldTwo = [b1, b2, b3, ...]

fieldThree = [c1, c2, c3, ...]

You can do this using list comprehension: "Make a list by taking the kth value from each element of myList". That would look like this:

fieldOne = [i[0] for i in myList]    # k = 0

fieldTwo = [i[1] for i in myList]    # k = 1

fieldThree = [i[2] for i in myList]  # k = 2

Of course, if your tuples are of variable length this gets much more complicated - but these are the fundamentals, and you can play around with them as needed to solve your problem.

answered Nov 20 '18 at 3:49

Green Cloak Guy

2,5131720

add a comment |

You have a list of 3-tuples:

myList = [

    (a1, b1, c1),

    (a2, b2, c2),

    (a3, b3, c3),

    ...

]

You want to refactor this list into three variables:

fieldOne = [a1, a2, a3, ...]

fieldTwo = [b1, b2, b3, ...]

fieldThree = [c1, c2, c3, ...]

You can do this using list comprehension: "Make a list by taking the kth value from each element of myList". That would look like this:

fieldOne = [i[0] for i in myList]    # k = 0

fieldTwo = [i[1] for i in myList]    # k = 1

fieldThree = [i[2] for i in myList]  # k = 2

Of course, if your tuples are of variable length this gets much more complicated - but these are the fundamentals, and you can play around with them as needed to solve your problem.

answered Nov 20 '18 at 3:49

Green Cloak Guy

2,5131720

You have a list of 3-tuples:

myList = [

    (a1, b1, c1),

    (a2, b2, c2),

    (a3, b3, c3),

    ...

]

You want to refactor this list into three variables:

fieldOne = [a1, a2, a3, ...]

fieldTwo = [b1, b2, b3, ...]

fieldThree = [c1, c2, c3, ...]

You can do this using list comprehension: "Make a list by taking the kth value from each element of myList". That would look like this:

fieldOne = [i[0] for i in myList]    # k = 0

fieldTwo = [i[1] for i in myList]    # k = 1

fieldThree = [i[2] for i in myList]  # k = 2

Of course, if your tuples are of variable length this gets much more complicated - but these are the fundamentals, and you can play around with them as needed to solve your problem.

answered Nov 20 '18 at 3:49

Green Cloak Guy

2,5131720

answered Nov 20 '18 at 3:49

Green Cloak Guy

2,5131720

answered Nov 20 '18 at 3:49

Green Cloak Guy

2,5131720

answered Nov 20 '18 at 3:49

Green Cloak Guy

2,5131720

add a comment |

You could also zip() your results and store them in a dictionary:

>>> myList = [('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106')]

>>> fields = 'fieldOne', 'fieldTwo', 'fieldThree'

>>> dict(zip(fields, zip(*myList)))

{'fieldOne': ('17.0.1.34', '203.84.210.248', '202.73.40.45'), 'fieldTwo': ('17.1.182.21', '27.111.228.3', '27.111.228.6'), 'fieldThree': ('714', '10310', '18106')}

Then you can just reference the 'fieldOne', 'fieldTwo', 'fieldThree' keys from this dictionary and store it in your database.

answered Nov 20 '18 at 3:57

RoadRunner

11.1k31340

add a comment |

You could also zip() your results and store them in a dictionary:

>>> myList = [('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106')]

>>> fields = 'fieldOne', 'fieldTwo', 'fieldThree'

>>> dict(zip(fields, zip(*myList)))

{'fieldOne': ('17.0.1.34', '203.84.210.248', '202.73.40.45'), 'fieldTwo': ('17.1.182.21', '27.111.228.3', '27.111.228.6'), 'fieldThree': ('714', '10310', '18106')}

Then you can just reference the 'fieldOne', 'fieldTwo', 'fieldThree' keys from this dictionary and store it in your database.

answered Nov 20 '18 at 3:57

RoadRunner

11.1k31340

add a comment |

You could also zip() your results and store them in a dictionary:

>>> myList = [('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106')]

>>> fields = 'fieldOne', 'fieldTwo', 'fieldThree'

>>> dict(zip(fields, zip(*myList)))

{'fieldOne': ('17.0.1.34', '203.84.210.248', '202.73.40.45'), 'fieldTwo': ('17.1.182.21', '27.111.228.3', '27.111.228.6'), 'fieldThree': ('714', '10310', '18106')}

Then you can just reference the 'fieldOne', 'fieldTwo', 'fieldThree' keys from this dictionary and store it in your database.

answered Nov 20 '18 at 3:57

RoadRunner

11.1k31340

You could also zip() your results and store them in a dictionary:

>>> myList = [('17.0.1.34', '17.1.182.21', '714'),('203.84.210.248', '27.111.228.3', '10310'),('202.73.40.45', '27.111.228.6', '18106')]

>>> fields = 'fieldOne', 'fieldTwo', 'fieldThree'

>>> dict(zip(fields, zip(*myList)))

{'fieldOne': ('17.0.1.34', '203.84.210.248', '202.73.40.45'), 'fieldTwo': ('17.1.182.21', '27.111.228.3', '27.111.228.6'), 'fieldThree': ('714', '10310', '18106')}

Then you can just reference the 'fieldOne', 'fieldTwo', 'fieldThree' keys from this dictionary and store it in your database.

answered Nov 20 '18 at 3:57

RoadRunner

11.1k31340

answered Nov 20 '18 at 3:57

RoadRunner

11.1k31340

answered Nov 20 '18 at 3:57

RoadRunner

11.1k31340

answered Nov 20 '18 at 3:57

RoadRunner

11.1k31340

add a comment |

I would suggest using the following way so you can use it multiple times.

def create_field(myList):

    fieldOne = [myList[i][0] for i in range(len(myList))]

    fieldTwo = [myList[i][1] for i in range(len(myList))]

    fieldThree = [myList[i][2] for i in range(len(myList))]

    return fieldOne, fieldTwo, fieldThree

create_field(myList)

answered Nov 20 '18 at 4:05

hemanta

178

add a comment |

I would suggest using the following way so you can use it multiple times.

def create_field(myList):

    fieldOne = [myList[i][0] for i in range(len(myList))]

    fieldTwo = [myList[i][1] for i in range(len(myList))]

    fieldThree = [myList[i][2] for i in range(len(myList))]

    return fieldOne, fieldTwo, fieldThree

create_field(myList)

answered Nov 20 '18 at 4:05

hemanta

178

add a comment |

I would suggest using the following way so you can use it multiple times.

def create_field(myList):

    fieldOne = [myList[i][0] for i in range(len(myList))]

    fieldTwo = [myList[i][1] for i in range(len(myList))]

    fieldThree = [myList[i][2] for i in range(len(myList))]

    return fieldOne, fieldTwo, fieldThree

create_field(myList)

answered Nov 20 '18 at 4:05

hemanta

178

I would suggest using the following way so you can use it multiple times.

def create_field(myList):

    fieldOne = [myList[i][0] for i in range(len(myList))]

    fieldTwo = [myList[i][1] for i in range(len(myList))]

    fieldThree = [myList[i][2] for i in range(len(myList))]

    return fieldOne, fieldTwo, fieldThree

create_field(myList)

answered Nov 20 '18 at 4:05

hemanta

178

answered Nov 20 '18 at 4:05

hemanta

178

answered Nov 20 '18 at 4:05

hemanta

178

answered Nov 20 '18 at 4:05

hemanta

178

add a comment |

Given your data, here is a possible solution:

# Initialize each field list

fieldOne = 

fieldTwo = 

fieldThree = 



# fetch each **item** in **myList**, we extract the component corresponding

# append it to the appropriate field

for item in myList:

    fieldOne.append(item[0])

    fieldTwo.append(item[1])

    fieldThree.append(item[2])

edited Nov 20 '18 at 4:25

answered Nov 20 '18 at 4:00

eapetcho

42927

add a comment |

Given your data, here is a possible solution:

# Initialize each field list

fieldOne = 

fieldTwo = 

fieldThree = 



# fetch each **item** in **myList**, we extract the component corresponding

# append it to the appropriate field

for item in myList:

    fieldOne.append(item[0])

    fieldTwo.append(item[1])

    fieldThree.append(item[2])

edited Nov 20 '18 at 4:25

answered Nov 20 '18 at 4:00

eapetcho

42927

add a comment |

Given your data, here is a possible solution:

# Initialize each field list

fieldOne = 

fieldTwo = 

fieldThree = 



# fetch each **item** in **myList**, we extract the component corresponding

# append it to the appropriate field

for item in myList:

    fieldOne.append(item[0])

    fieldTwo.append(item[1])

    fieldThree.append(item[2])

edited Nov 20 '18 at 4:25

answered Nov 20 '18 at 4:00

eapetcho

42927

Given your data, here is a possible solution:

# Initialize each field list

fieldOne = 

fieldTwo = 

fieldThree = 



# fetch each **item** in **myList**, we extract the component corresponding

# append it to the appropriate field

for item in myList:

    fieldOne.append(item[0])

    fieldTwo.append(item[1])

    fieldThree.append(item[2])

edited Nov 20 '18 at 4:25

answered Nov 20 '18 at 4:00

eapetcho

42927

edited Nov 20 '18 at 4:25

answered Nov 20 '18 at 4:00

eapetcho

42927

answered Nov 20 '18 at 4:00

eapetcho

42927

answered Nov 20 '18 at 4:00

eapetcho

42927

add a comment |

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