快速遍历整个python字典的方法

Question

I have non-ordered data that sometimes I want to analyse by looking at all the entries and some other time I want to pick just one entry.

p1   x1 x2 x3 x4
p2   x1 x2 x3 x4
p33  x1 x2 x3 x4
p3   x1 x2 x3 x4
p4   x1 x2 x3 x4

Dictionary seems a nice format to store the data, as it is not sorted, and if I want to get p33 , which might be anywhere in the table I can do that by dict["p33"] . This lookup will take some time, but I suppose is faster than looping on the whole data to find the line that I want (at least this is the advantage I have been advertised dict should buy me).

If I want to look at the whole data, eg counting how many times x3 is zero, I should loop on all the lines and doing it by a for loop of the type for item in dict.keys(): is too slow. I have the impression that getting the keys and then doing dict[item] make a lot of useless lookup, because for each item it has to find it in the dictionary, whereas for my goal would be good enough to read serially "as if it were a list".

So I was wondering if there is a faster way to loop on all the entries of the dictionary.

Thanks

Answer 1

If its possible use numpy/pandas...

For me Python is only for High Level Programming and Low Level is C++... So if possible use existing c++ functions which are in numpy pandas or other libs..

Check it out...

>>> import numpy as np, pandas as pd
>>> p1 = np.arange(10)
>>> dct = dict(
... p1 = np.arange(10),
... p2 = np.ones(10),
... p3 = np.zeros(10),
... p33 = np.ones(10)*10,
... p4 = np.linspace(0,1,10))
>>>
>>> dct
{'p2': array([ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.]), 'p33': array([ 10.,  10.,  10.,  10.,  10.,
10.,  10.,  10.,  10.,  10.]), 'p1': array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]), 'p4': array([ 0.        ,  0.11111111,  0.22222222,  0.33333333,  0.44444444,
        0.55555556,  0.66666667,  0.77777778,  0.88888889,  1.        ]), 'p3': array([ 0.,  0.,  0.,  0.,  0.,
  0.,  0.,  0.,  0.,  0.])}
>>> from pprint import pprint as pr
>>> pr(dct)
{'p1': array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]),
 'p2': array([ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.]),
 'p3': array([ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]),
 'p33': array([ 10.,  10.,  10.,  10.,  10.,  10.,  10.,  10.,  10.,  10.]),
 'p4': array([ 0.        ,  0.11111111,  0.22222222,  0.33333333,  0.44444444,
        0.55555556,  0.66666667,  0.77777778,  0.88888889,  1.        ])}
>>> df = pd.DataFrame(dct)
>>> df
   p1   p2   p3   p33        p4
0   0  1.0  0.0  10.0  0.000000
1   1  1.0  0.0  10.0  0.111111
2   2  1.0  0.0  10.0  0.222222
3   3  1.0  0.0  10.0  0.333333
4   4  1.0  0.0  10.0  0.444444
5   5  1.0  0.0  10.0  0.555556
6   6  1.0  0.0  10.0  0.666667
7   7  1.0  0.0  10.0  0.777778
8   8  1.0  0.0  10.0  0.888889
9   9  1.0  0.0  10.0  1.000000
>>> df.T
        0          1          2          3          4          5          6  \
p1    0.0   1.000000   2.000000   3.000000   4.000000   5.000000   6.000000
p2    1.0   1.000000   1.000000   1.000000   1.000000   1.000000   1.000000
p3    0.0   0.000000   0.000000   0.000000   0.000000   0.000000   0.000000
p33  10.0  10.000000  10.000000  10.000000  10.000000  10.000000  10.000000
p4    0.0   0.111111   0.222222   0.333333   0.444444   0.555556   0.666667

             7          8     9
p1    7.000000   8.000000   9.0
p2    1.000000   1.000000   1.0
p3    0.000000   0.000000   0.0
p33  10.000000  10.000000  10.0
p4    0.777778   0.888889   1.0
>>> df = df.T
>>> df.columns = ['x%d'%(n+1) for n in df.columns.values]
>>> df
       x1         x2         x3         x4         x5         x6         x7  \
p1    0.0   1.000000   2.000000   3.000000   4.000000   5.000000   6.000000
p2    1.0   1.000000   1.000000   1.000000   1.000000   1.000000   1.000000
p3    0.0   0.000000   0.000000   0.000000   0.000000   0.000000   0.000000
p33  10.0  10.000000  10.000000  10.000000  10.000000  10.000000  10.000000
p4    0.0   0.111111   0.222222   0.333333   0.444444   0.555556   0.666667

            x8         x9   x10
p1    7.000000   8.000000   9.0
p2    1.000000   1.000000   1.0
p3    0.000000   0.000000   0.0
p33  10.000000  10.000000  10.0
p4    0.777778   0.888889   1.0
>>> df.x3
p1      2.000000
p2      1.000000
p3      0.000000
p33    10.000000
p4      0.222222
Name: x3, dtype: float64
>>> df.x3 == 0
p1     False
p2     False
p3      True
p33    False
p4     False
Name: x3, dtype: bool
>>> np.sum(df.x3 == 0)
1
>>>