45

I have a dictionary that I need to convert to a NumPy structured array. I'm using the arcpy function NumPyArraytoTable, so a NumPy structured array is the only data format that will work.

Based on this thread: Writing to numpy array from dictionary and this thread: How to convert Python dictionary object to numpy array

I've tried this:

result = {0: 1.1181753789488595, 1: 0.5566080288678394, 2: 0.4718269778030734, 3: 0.48716683119447185, 4: 1.0, 5: 0.1395076201641266, 6: 0.20941558441558442}

names = ['id','data']
formats = ['f8','f8']
dtype = dict(names = names, formats=formats)
array=numpy.array([[key,val] for (key,val) in result.iteritems()],dtype)

But I keep getting expected a readable buffer object

The method below works, but is stupid and obviously won't work for real data. I know there is a more graceful approach, I just can't figure it out.

totable = numpy.array([[key,val] for (key,val) in result.iteritems()])
array=numpy.array([(totable[0,0],totable[0,1]),(totable[1,0],totable[1,1])],dtype)
4

5 回答 5

70

You could use np.array(list(result.items()), dtype=dtype):

import numpy as np
result = {0: 1.1181753789488595, 1: 0.5566080288678394, 2: 0.4718269778030734, 3: 0.48716683119447185, 4: 1.0, 5: 0.1395076201641266, 6: 0.20941558441558442}

names = ['id','data']
formats = ['f8','f8']
dtype = dict(names = names, formats=formats)
array = np.array(list(result.items()), dtype=dtype)

print(repr(array))

yields

array([(0.0, 1.1181753789488595), (1.0, 0.5566080288678394),
       (2.0, 0.4718269778030734), (3.0, 0.48716683119447185), (4.0, 1.0),
       (5.0, 0.1395076201641266), (6.0, 0.20941558441558442)], 
      dtype=[('id', '<f8'), ('data', '<f8')])

If you don't want to create the intermediate list of tuples, list(result.items()), then you could instead use np.fromiter:

In Python2:

array = np.fromiter(result.iteritems(), dtype=dtype, count=len(result))

In Python3:

array = np.fromiter(result.items(), dtype=dtype, count=len(result))

Why using the list [key,val] does not work:

By the way, your attempt, 

numpy.array([[key,val] for (key,val) in result.iteritems()],dtype)

was very close to working. If you change the list [key, val] to the tuple (key, val), then it would have worked. Of course, 

numpy.array([(key,val) for (key,val) in result.iteritems()], dtype)

is the same thing as 

numpy.array(result.items(), dtype)

in Python2, or

numpy.array(list(result.items()), dtype)

in Python3.

np.array treats lists differently than tuples: Robert Kern explains:

As a rule, tuples are considered "scalar" records and lists are recursed upon. This rule helps numpy.array() figure out which sequences are records and which are other sequences to be recursed upon; i.e. which sequences create another dimension and which are the atomic elements.

Since (0.0, 1.1181753789488595) is considered one of those atomic elements, it should be a tuple, not a list.

于 2013-03-22T21:04:15.840 回答
5

Even more simple if you accept using pandas :

import pandas
result = {0: 1.1181753789488595, 1: 0.5566080288678394, 2: 0.4718269778030734, 3: 0.48716683119447185, 4: 1.0, 5: 0.1395076201641266, 6: 0.20941558441558442}
df = pandas.DataFrame(result, index=[0])
print df

gives :

          0         1         2         3  4         5         6
0  1.118175  0.556608  0.471827  0.487167  1  0.139508  0.209416
于 2017-04-03T14:30:06.230 回答
3

Let me propose an improved method when the values of the dictionnary are lists with the same lenght :

import numpy

def dctToNdarray (dd, szFormat = 'f8'):
    '''
    Convert a 'rectangular' dictionnary to numpy NdArray
    entry 
        dd : dictionnary (same len of list 
    retrun
        data : numpy NdArray 
    '''
    names = dd.keys()
    firstKey = dd.keys()[0]
    formats = [szFormat]*len(names)
    dtype = dict(names = names, formats=formats)
    values = [tuple(dd[k][0] for k in dd.keys())]
    data = numpy.array(values, dtype=dtype)
    for i in range(1,len(dd[firstKey])) :
        values = [tuple(dd[k][i] for k in dd.keys())]
        data_tmp = numpy.array(values, dtype=dtype)
        data = numpy.concatenate((data,data_tmp))
    return data

dd = {'a':[1,2.05,25.48],'b':[2,1.07,9],'c':[3,3.01,6.14]}
data = dctToNdarray(dd)
print data.dtype.names
print data
于 2015-02-18T08:35:12.637 回答
2

I would prefer storing keys and values on separate arrays. This i often more practical. Structures of arrays are perfect replacement to array of structures. As most of the time you have to process only a subset of your data (in this cases keys or values, operation only with only one of the two arrays would be more efficient than operating with half of the two arrays together.

But in case this way is not possible, I would suggest to use arrays sorted by column instead of by row. In this way you would have the same benefit as having two arrays, but packed only in one.

import numpy as np
result = {0: 1.1181753789488595, 1: 0.5566080288678394, 2: 0.4718269778030734, 3: 0.48716683119447185, 4: 1.0, 5: 0.1395076201641266, 6: 0.20941558441558442}

names = 0
values = 1
array = np.empty(shape=(2, len(result)), dtype=float)
array[names] = result.keys()
array[values] = result.values()

But my favorite is this (simpler):

import numpy as np
result = {0: 1.1181753789488595, 1: 0.5566080288678394, 2: 0.4718269778030734, 3: 0.48716683119447185, 4: 1.0, 5: 0.1395076201641266, 6: 0.20941558441558442}

arrays = {'names': np.array(result.keys(), dtype=float),
          'values': np.array(result.values(), dtype=float)}
于 2015-03-07T01:51:50.463 回答
1

Similarly to the approved answer. If you want to create an array from dictionary keys:

np.array( tuple(dict.keys()) )

If you want to create an array from dictionary values:

np.array( tuple(dict.values()) )
于 2019-08-17T13:57:46.103 回答