In [9]: pd.Series(df.Letter.values,index=df.Position).to_dict()
Out[9]: {1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e'}

Speed comparion (using Wouter's method)

In [6]: df = pd.DataFrame(randint(0,10,10000).reshape(5000,2),columns=list('AB'))

In [7]: %timeit dict(zip(df.A,df.B))
1000 loops, best of 3: 1.27 ms per loop

In [8]: %timeit pd.Series(df.A.values,index=df.B).to_dict()
1000 loops, best of 3: 987 us per loop

I found a faster way to solve the problem, at least on realistically large datasets using: df.set_index(KEY).to_dict()[VALUE]

Proof on 50,000 rows:

df = pd.DataFrame(np.random.randint(32, 120, 100000).reshape(50000,2),columns=list('AB'))
df['A'] = df['A'].apply(chr)

%timeit dict(zip(df.A,df.B))
%timeit pd.Series(df.A.values,index=df.B).to_dict()
%timeit df.set_index('A').to_dict()['B']

Output:

100 loops, best of 3: 7.04 ms per loop  # WouterOvermeire
100 loops, best of 3: 9.83 ms per loop  # Jeff
100 loops, best of 3: 4.28 ms per loop  # Kikohs (me)

In Python 3.6 the fastest way is still the WouterOvermeire one. Kikohs' proposal is slower than the other two options.

import timeit

setup = '''
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randint(32, 120, 100000).reshape(50000,2),columns=list('AB'))
df['A'] = df['A'].apply(chr)
'''

timeit.Timer('dict(zip(df.A,df.B))', setup=setup).repeat(7,500)
timeit.Timer('pd.Series(df.A.values,index=df.B).to_dict()', setup=setup).repeat(7,500)
timeit.Timer('df.set_index("A").to_dict()["B"]', setup=setup).repeat(7,500)

Results:

1.1214002349999777 s  # WouterOvermeire
1.1922008498571748 s  # Jeff
1.7034366211428602 s  # Kikohs

TL;DR

>>> import pandas as pd
>>> df = pd.DataFrame({'Position':[1,2,3,4,5], 'Letter':['a', 'b', 'c', 'd', 'e']})
>>> dict(sorted(df.values.tolist())) # Sort of sorted... 
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
>>> from collections import OrderedDict
>>> OrderedDict(df.values.tolist())
OrderedDict([('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', 5)])

In Long

Explaining solution: dict(sorted(df.values.tolist()))

Given:

df = pd.DataFrame({'Position':[1,2,3,4,5], 'Letter':['a', 'b', 'c', 'd', 'e']})

[out]:

 Letter Position
0   a   1
1   b   2
2   c   3
3   d   4
4   e   5

Try:

# Get the values out to a 2-D numpy array, 
df.values

[out]:

array([['a', 1],
       ['b', 2],
       ['c', 3],
       ['d', 4],
       ['e', 5]], dtype=object)

Then optionally:

# Dump it into a list so that you can sort it using `sorted()`
sorted(df.values.tolist()) # Sort by key

Or:

# Sort by value:
from operator import itemgetter
sorted(df.values.tolist(), key=itemgetter(1))

[out]:

[['a', 1], ['b', 2], ['c', 3], ['d', 4], ['e', 5]]

Lastly, cast the list of list of 2 elements into a dict.

dict(sorted(df.values.tolist())) 

[out]:

{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}

Related

Answering @sbradbio comment:

If there are multiple values for a specific key and you would like to keep all of them, it's the not the most efficient but the most intuitive way is:

from collections import defaultdict
import pandas as pd

multivalue_dict = defaultdict(list)

df = pd.DataFrame({'Position':[1,2,4,4,4], 'Letter':['a', 'b', 'd', 'e', 'f']})

for idx,row in df.iterrows():
    multivalue_dict[row['Position']].append(row['Letter'])

[out]:

>>> print(multivalue_dict)
defaultdict(list, {1: ['a'], 2: ['b'], 4: ['d', 'e', 'f']})