# 创建两个熊猫数据框列的字典的最有效方法是什么？

## 问题：创建两个熊猫数据框列的字典的最有效方法是什么？

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

What is the most efficient way to organise the following pandas Dataframe:

data =

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

into a dictionary like `alphabet[1 : 'a', 2 : 'b', 3 : 'c', 4 : 'd', 5 : 'e']`?

## 回答 0

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

``````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``````
``````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
``````

## 回答 1

50,000行的证明：

``````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']``````

``````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)``````

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)
``````

## 回答 2

Python 3.6中，最快的方法仍然是WouterOvermeire。Kikohs的提议比其他两个方案要慢。

``````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)``````

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

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)])``````

# 在长

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

[出]：

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

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

[出]：

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

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

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

[出]：

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

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

[出]：

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

# 有关

``````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'])``````

[出]：

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

# 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

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']})
``````