How to retieve value from a dictionary based on key ?

up vote
-1
down vote

favorite

I am stuck at this minor problem for quite some time and cannot understand why.

lets say I have a list :

test = ['1', '2', '3']

I convert this into dictionary using

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

Now when I access key based on value like this

print (a.get('1'))

it gives me None. Any suggestions in this regard would be helpful.

asked Nov 11 at 15:09

Muss

296

Try print (test_dict[1]) #=> 2
– iGian
Nov 11 at 15:12

Dictionaries are a unidirectional mapping of keys to values.
– timgeb
Nov 11 at 15:14

I want to print the key not value. This would give 2 which the is the value of 1. I want to do it the other way round
– Muss
Nov 11 at 15:15

add a comment |

up vote
-1
down vote

favorite

I am stuck at this minor problem for quite some time and cannot understand why.

lets say I have a list :

test = ['1', '2', '3']

I convert this into dictionary using

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

Now when I access key based on value like this

print (a.get('1'))

it gives me None. Any suggestions in this regard would be helpful.

asked Nov 11 at 15:09

Muss

296

Try print (test_dict[1]) #=> 2
– iGian
Nov 11 at 15:12

Dictionaries are a unidirectional mapping of keys to values.
– timgeb
Nov 11 at 15:14

I want to print the key not value. This would give 2 which the is the value of 1. I want to do it the other way round
– Muss
Nov 11 at 15:15

add a comment |

up vote
-1
down vote

favorite

I am stuck at this minor problem for quite some time and cannot understand why.

lets say I have a list :

test = ['1', '2', '3']

I convert this into dictionary using

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

Now when I access key based on value like this

print (a.get('1'))

it gives me None. Any suggestions in this regard would be helpful.

asked Nov 11 at 15:09

Muss

296

I am stuck at this minor problem for quite some time and cannot understand why.

lets say I have a list :

test = ['1', '2', '3']

I convert this into dictionary using

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

Now when I access key based on value like this

print (a.get('1'))

it gives me None. Any suggestions in this regard would be helpful.

python

asked Nov 11 at 15:09

Muss

296

asked Nov 11 at 15:09

Muss

296

asked Nov 11 at 15:09

Muss

296

asked Nov 11 at 15:09

Muss

296

asked Nov 11 at 15:09

Muss

296

Try print (test_dict[1]) #=> 2
– iGian
Nov 11 at 15:12

Dictionaries are a unidirectional mapping of keys to values.
– timgeb
Nov 11 at 15:14

I want to print the key not value. This would give 2 which the is the value of 1. I want to do it the other way round
– Muss
Nov 11 at 15:15

add a comment |

Try print (test_dict[1]) #=> 2
– iGian
Nov 11 at 15:12

Dictionaries are a unidirectional mapping of keys to values.
– timgeb
Nov 11 at 15:14

I want to print the key not value. This would give 2 which the is the value of 1. I want to do it the other way round
– Muss
Nov 11 at 15:15

Try print (test_dict[1]) #=> 2
– iGian
Nov 11 at 15:12

Dictionaries are a unidirectional mapping of keys to values.
– timgeb
Nov 11 at 15:14

I want to print the key not value. This would give 2 which the is the value of 1. I want to do it the other way round
– Muss
Nov 11 at 15:15

add a comment |

1 Answer
1

active

oldest

votes

up vote
1
down vote

Its a problem of type

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

The keys will be of integer type and not string type.

Try this out instead

print (a.get(1))

Edit:
To get the keys, you can flip the dictionary creation

test_dict =  { test[i] : i for i in range(0, len(test))}



print (a.get(1))

0

edited Nov 11 at 15:24

answered Nov 11 at 15:14

Sharath

114

It gives the value of 1. I need to get key instead like if I do test_dict[1] it should give me 0
– Muss
Nov 11 at 15:20

Flip the dictionary creation around then. Check the answer I have edited it.
– Sharath
Nov 11 at 15:23

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

Its a problem of type

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

The keys will be of integer type and not string type.

Try this out instead

print (a.get(1))

Edit:
To get the keys, you can flip the dictionary creation

test_dict =  { test[i] : i for i in range(0, len(test))}



print (a.get(1))

0

edited Nov 11 at 15:24

answered Nov 11 at 15:14

Sharath

114

It gives the value of 1. I need to get key instead like if I do test_dict[1] it should give me 0
– Muss
Nov 11 at 15:20

Flip the dictionary creation around then. Check the answer I have edited it.
– Sharath
Nov 11 at 15:23

add a comment |

up vote
1
down vote

Its a problem of type

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

The keys will be of integer type and not string type.

Try this out instead

print (a.get(1))

Edit:
To get the keys, you can flip the dictionary creation

test_dict =  { test[i] : i for i in range(0, len(test))}



print (a.get(1))

0

edited Nov 11 at 15:24

answered Nov 11 at 15:14

Sharath

114

It gives the value of 1. I need to get key instead like if I do test_dict[1] it should give me 0
– Muss
Nov 11 at 15:20

Flip the dictionary creation around then. Check the answer I have edited it.
– Sharath
Nov 11 at 15:23

add a comment |

up vote
1
down vote

Its a problem of type

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

The keys will be of integer type and not string type.

Try this out instead

print (a.get(1))

Edit:
To get the keys, you can flip the dictionary creation

test_dict =  { test[i] : i for i in range(0, len(test))}



print (a.get(1))

0

edited Nov 11 at 15:24

answered Nov 11 at 15:14

Sharath

114

Its a problem of type

test_dict =  { i : test[i] for i in range(0, len(test))}



{0: '1', 1: '2', 2: '3'}

The keys will be of integer type and not string type.

Try this out instead

print (a.get(1))

Edit:
To get the keys, you can flip the dictionary creation

test_dict =  { test[i] : i for i in range(0, len(test))}



print (a.get(1))

0

edited Nov 11 at 15:24

answered Nov 11 at 15:14

Sharath

114

edited Nov 11 at 15:24

answered Nov 11 at 15:14

Sharath

114

answered Nov 11 at 15:14

Sharath

114

answered Nov 11 at 15:14

Sharath

114

It gives the value of 1. I need to get key instead like if I do test_dict[1] it should give me 0
– Muss
Nov 11 at 15:20

Flip the dictionary creation around then. Check the answer I have edited it.
– Sharath
Nov 11 at 15:23

add a comment |

It gives the value of 1. I need to get key instead like if I do test_dict[1] it should give me 0
– Muss
Nov 11 at 15:20

Flip the dictionary creation around then. Check the answer I have edited it.
– Sharath
Nov 11 at 15:23

It gives the value of 1. I need to get key instead like if I do test_dict[1] it should give me 0
– Muss
Nov 11 at 15:20

Flip the dictionary creation around then. Check the answer I have edited it.
– Sharath
Nov 11 at 15:23

add a comment |

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