I have a pandas dataframe with a multiindex. Unfortunately one of the indices gives years as a string
e.g. '2010', '2011'
how do I convert these to integers?
More concretely
MultiIndex(levels=[[u'2010', u'2011'], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]],
labels=[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, , ...]], names=[u'Year', u'Month'])
.
df_cbs_prelim_total.index.set_levels(df_cbs_prelim_total.index.get_level_values(0).astype('int'))
seems to do it, but not inplace. Any proper way of changing them?
Cheers,
Mike
Will probably be cleaner to do this before you assign it as index (as #EdChum points out), but when you already have it as index, you can indeed use set_levels to alter one of the labels of a level of your multi-index. A bit cleaner as your code (you can use index.levels[..]):
In [165]: idx = pd.MultiIndex.from_product([[1,2,3], ['2011','2012','2013']])
In [166]: idx
Out[166]:
MultiIndex(levels=[[1, 2, 3], [u'2011', u'2012', u'2013']],
labels=[[0, 0, 0, 1, 1, 1, 2, 2, 2], [0, 1, 2, 0, 1, 2, 0, 1, 2]])
In [167]: idx.levels[1]
Out[167]: Index([u'2011', u'2012', u'2013'], dtype='object')
In [168]: idx = idx.set_levels(idx.levels[1].astype(int), level=1)
In [169]: idx
Out[169]:
MultiIndex(levels=[[1, 2, 3], [2011, 2012, 2013]],
labels=[[0, 0, 0, 1, 1, 1, 2, 2, 2], [0, 1, 2, 0, 1, 2, 0, 1, 2]])
You have to reassign it to save the changes (as is done above, in your case this would be df_cbs_prelim_total.index = df_cbs_prelim_total.index.set_levels(...))
Related
how to split numpy array by step?
Example:
I have array:
[3, 0, 5, 0, 7, 0, 3, 1]
I want to spit like this:
[3, 5, 6, 3]
[0, 0, 0, 1]
Or a more understandable example:
['a1', 'a2', 'b1', 'b2'] -- > ['a1', 'b1'] and ['a2', 'b2']
You can do this with array slicing.
arr = np.array([3, 0, 5, 0, 7, 0, 3, 1])
A = arr[::2]
B = arr[1::2]
see docs on slices here
Say I have a list of Indices:
np.array([1, 3, 2, 4])
How do I create the following matrix, where all elements left to the index are ones and right to the index zeros?
[[1, 1, 0, 0, 0, 0],
[1, 1, 1, 1, 0, 0],
[1, 1, 1, 0, 0, 0],
[1, 1, 1, 1, 1, 0]]
1*(np.arange( 6 ) <= arr[:,None])
# array([[1, 1, 0, 0, 0, 0],
# [1, 1, 1, 1, 0, 0],
# [1, 1, 1, 0, 0, 0],
# [1, 1, 1, 1, 1, 0]])
This broadcasts the array of 6 elements across the rows and the array of indices across the columns. The 1* converts boolean to int.
I am trying to find a way of Counting zeros in a rolling using numpy array ?
Using pandas I can get it using:
df['demand'].apply(lambda x: (x == 0).rolling(7).sum()).fillna(0))
or
df['demand'].transform(lambda x: x.rolling(7).apply(lambda x: 7 - np.count _nonzero(x))).fillna(0)
In numpy, using the code from Here
def rolling_window(a, window_size):
shape = (a.shape[0] - window_size + 1, window_size) + a.shape[1:]
print(shape)
strides = (a.strides[0],) + a.strides
return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
arr = np.asarray([10, 20, 30, 5, 6, 0, 0, 0])
np.count_nonzero(rolling_window(arr==0, 7), axis=1)
Output:
array([2, 3])
However, I need the first 6 NaNs as well, and fill it with zeros:
Expected output:
array([0, 0, 0, 0, 0, 0, 2, 3])
Think an efficient one would be with 1D convolution -
def sum_occurences_windowed(arr, W):
K = np.ones(W, dtype=int)
out = np.convolve(arr==0,K)[:len(arr)]
out[:W-1] = 0
return out
Sample run -
In [42]: arr
Out[42]: array([10, 20, 30, 5, 6, 0, 0, 0])
In [43]: sum_occurences_windowed(arr,W=7)
Out[43]: array([0, 0, 0, 0, 0, 0, 2, 3])
Timings on varying length arrays and window of 7
Including count_rolling from #Quang Hoang's post.
Using benchit package (few benchmarking tools packaged together; disclaimer: I am its author) to benchmark proposed solutions.
import benchit
funcs = [sum_occurences_windowed, count_rolling]
in_ = {n:(np.random.randint(0,5,(n)),7) for n in [10,20,50,100,200,500,1000,2000,5000]}
t = benchit.timings(funcs, in_, multivar=True, input_name='Length')
t.plot(logx=True, save='timings.png')
Extending to generic n-dim arrays
from scipy.ndimage.filters import convolve1d
def sum_occurences_windowed_ndim(arr, W, axis=-1):
K = np.ones(W, dtype=int)
out = convolve1d((arr==0).astype(int),K,axis=axis,origin=-(W//2))
out.swapaxes(axis,0)[:W-1] = 0
return out
So, on a 2D array, for counting along each row, use axis=1 and for cols, axis=0 and so on.
Sample run -
In [155]: np.random.seed(0)
In [156]: a = np.random.randint(0,3,(3,10))
In [157]: a
Out[157]:
array([[0, 1, 0, 1, 1, 2, 0, 2, 0, 0],
[0, 2, 1, 2, 2, 0, 1, 1, 1, 1],
[0, 1, 0, 0, 1, 2, 0, 2, 0, 1]])
In [158]: sum_occurences_windowed_ndim(a, W=7)
Out[158]:
array([[0, 0, 0, 0, 0, 0, 3, 2, 3, 3],
[0, 0, 0, 0, 0, 0, 2, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 4, 3, 4, 3]])
# Verify with earlier 1D solution
In [159]: np.vstack([sum_occurences_windowed(i,7) for i in a])
Out[159]:
array([[0, 0, 0, 0, 0, 0, 3, 2, 3, 3],
[0, 0, 0, 0, 0, 0, 2, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 4, 3, 4, 3]])
Let's test out our original 1D input array -
In [187]: arr
Out[187]: array([10, 20, 30, 5, 6, 0, 0, 0])
In [188]: sum_occurences_windowed_ndim(arr, W=7)
Out[188]: array([0, 0, 0, 0, 0, 0, 2, 3])
I would modify the function as follow:
def count_rolling(a, window_size):
shape = (a.shape[0] - window_size + 1, window_size) + a.shape[1:]
strides = (a.strides[0],) + a.strides
rolling = np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
out = np.zeros_like(a)
out[window_size-1:] = (rolling == 0).sum(1)
return out
arr = np.asarray([10, 20, 30, 5, 6, 0, 0, 0])
count_rolling(arr,7)
Output:
array([0, 0, 0, 0, 0, 0, 2, 3])
How to place a list of numbers in to a 2D numpy array, where the second dimension of the array is equal to the number of digits of the largest number of that list? I also want the elements that don't belong to the original number to be zero in each row of the returning array.
Example:
From the list a = range(0,1001), how to get the numpy array of the below form:
[[0,0,0,0],
[0,0,0,1],
[0,0,0,2],
...
[0,9,9,8]
[0,9,9,9],
[1,0,0,0]]
Please note how the each number is placed in-place in a np.zeros((1000,4)) array at the end of the each row.
NB: A pythonic, vectorized implementation is expected
Broadcasting again!
def split_digits(a):
N = int(np.log10(np.max(a))+1) # No. of digits
r = 10**np.arange(N,-1,-1) # 10-powered range array
return (np.asarray(a)[:,None]%r[:-1])//r[1:]
Sample runs -
In [224]: a = range(0,1001)
In [225]: split_digits(a)
Out[225]:
array([[0, 0, 0, 0],
[0, 0, 0, 1],
[0, 0, 0, 2],
...,
[0, 9, 9, 8],
[0, 9, 9, 9],
[1, 0, 0, 0]])
In [229]: a = np.random.randint(0,1000000,(7))
In [230]: a
Out[230]: array([431921, 871855, 636144, 541186, 410562, 89356, 476258])
In [231]: split_digits(a)
Out[231]:
array([[4, 3, 1, 9, 2, 1],
[8, 7, 1, 8, 5, 5],
[6, 3, 6, 1, 4, 4],
[5, 4, 1, 1, 8, 6],
[4, 1, 0, 5, 6, 2],
[0, 8, 9, 3, 5, 6],
[4, 7, 6, 2, 5, 8]])
Another concept using pandas str
def pir(a):
z = int(np.log10(np.max(a)))
s = pd.Series(a.astype(str))
zfilled = s.str.zfill(z + 1).sum()
a_ = np.array(list(zfilled)).reshape(-1, z + 1)
return a_.astype(int)
Using #Divakar's random array
a = np.random.randint(0,1000000,(7))
array([ 57190, 29950, 392317, 592062, 460333, 639794, 983647])
pir(a)
array([[0, 5, 7, 1, 9, 0],
[0, 2, 9, 9, 5, 0],
[3, 9, 2, 3, 1, 7],
[5, 9, 2, 0, 6, 2],
[4, 6, 0, 3, 3, 3],
[6, 3, 9, 7, 9, 4],
[9, 8, 3, 6, 4, 7]])
Googled around a bit and couldn't seem to find anything on this.
Is there an option to access data in a pandas data frame using "not index"?
So something like
df_index = asdf = pandas.MultiIndex(levels=[
['2014-10-19', '2014-10-20', '2014-10-21', '2014-10-22', '2014-10-30'],
[u'after_work', u'all_day', u'breakfast', u'lunch', u'mid_evening']],
labels=[[0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 4, 4],
[4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 2, 0, 1, 3, 4]],
names=[u'start_date', u'time_group'])
And then I would like to be able to call the following to get everything not in df_index
df.ix[~df_index]
I know you can do it for logical indexing within pandas. Just curious if I could do it using an Index Object
you can use df.drop(df_index, errors="ignore").