Previously I used the following to calculate the ewma
dataset['26ema'] = pd.ewma(dataset['price'], span=26)
But, in the latest version of pandas pd.ewma has been removed. How to calculate using the new method dataframe.ewma?
dataset['26ema'] = dataset['price'].ewma(span=26)
This is giving an error 'AttributeError: 'Series' object has no attribute 'ewma'
Use Series.ewm:
dataset['price'].ewm(span=26)
See GH11603 for the relevant PR and mapping of the old API to new ones.
Minimal Code Example
s = pd.Series(range(5))
s.ewm(span=3).mean()
0 0.000000
1 0.666667
2 1.428571
3 2.266667
4 3.161290
dtype: float64
Related
I've been trying to optimise a bokeh server to calculate live stats by selected country on Covid19.
I found myself repeating a groupby function to calculate new columns and was wondering, having selected the groupby, if I could then apply it in a similar way to .agg() on multiple columns ?
For example:
dfall = pd.DataFrame(db("SELECT * FROM C19daily"))
dfall.set_index(['geoId', 'date'], drop=False, inplace=True)
dfall = dfall.sort_index(ascending=True)
dfall.head()
id date geoId cases deaths auid
geoId date
AD 2020-03-03 70119 2020-03-03 AD 1 0 AD03/03/2020
2020-03-14 70118 2020-03-14 AD 1 0 AD14/03/2020
2020-03-16 70117 2020-03-16 AD 3 0 AD16/03/2020
2020-03-17 70116 2020-03-17 AD 9 0 AD17/03/2020
2020-03-18 70115 2020-03-18 AD 0 0 AD18/03/2020
I need to create new columns based on 'cases' and 'deaths' and applying various functions like cumsum(). Currently I do this the long way
dfall['ccases'] = dfall.groupby(level=0)['cases'].cumsum()
dfall['dpc_cases'] = dfall.groupby(level=0)['cases'].pct_change(fill_method='pad', periods=7)
.....
dfall['cdeaths'] = dfall.groupby(level=0)['deaths'].cumsum()
dfall['dpc_deaths'] = dfall.groupby(level=0)['deaths'].pct_change(fill_method='pad', periods=7)
I tried to optimise the groupby call like this:-
with dfall.groupby(level=0) as gr:
gr = g['cases'].cumsum()...
But the error suggest the class doesn't support this
AttributeError: __enter__
I thought I could use .agg({}) and supply dictionary
g = dfall.groupby(level=0).agg({'cc' : 'cumsum', 'cd' : 'cumsum'})
but that produces another error
pandas.core.base.SpecificationError: nested renamer is not supported
I have plenty of other bits to optimise, I thought this python part would be the easiest and save a few ms!
Could anyone nudge me in the right direction?
To avoid repeating dfall.groupby(level=0) you can just save it in a variable:
gb = dfall.groupby(level=0)
gb_cases = gb['cases']
dfall['ccases'] = gb_cases.cumsum()
dfall['dpc_cases'] = gb_cases.pct_change(fill_method='pad', periods=7)
...
And to run multiple aggregations using a single expression, I think you can use named aggregation. But I have no clue whether it will be more performant or not. Either way, it's better to profile the code and improve the actual bottlenecks.
I have check the previous post link but it doesn't seems to work for my case:-
I have pre trained word2vec model:
import gensim
model = Word2Vec.load('w2v_model')
Now I have a pandas dataframe with keywords:
keyword
corruption
people
budget
cambodia
.......
......
All I want to add the vectors for each keyword in its corresponding columns but
when I use model['cambodia'] it throw me error as KeyError: "word 'cambodia' not in vocabulary"
so I have update the keyword as:
model.train(['cambodia'])
But this won't work out for me, when I use
model['cambodia']
it still giving an error as KeyError: "word 'cambodia' not in vocabulary". How to update new words into word2vec vocabulary so i can get its vectors? Expected output will be:-
keyword V1 V2 V3 V4 V5 V6
corruption 0.07397 0.290874 -0.170812 0.085428 -0.148551 0.38846
people ..............................................................
budget ...........................................................
You can initial the first vector as [0,0,...0]. And the word that not in vocabulary can set to 0.
keyword V1 V2 V3 V4 V5 V6
0 0 0 0 0 0 0
1 0.07397 0.290874 -0.170812 0.085428 -0.148551 0.38846
2 ..............................................................
3 ...........................................................
You can use two dicts to solve the problem.
word2id['corruption']=1
vec['corruption']=[0.07397 0.290874 -0.170812 0.085428 -0.148551 0.38846]
...
word2id['cambodia']=0
vec['cambodia']=[0 0 0 0 0 0]
I am currently building a binary classification model and have created an input file for svm-train (svm_input.txt). This input file has 453 lines, 4 No. features and 2 No. classes [0,1].
i.e
0 1:15.0 2:40.0 3:30.0 4:15.0
1 1:22.73 2:40.91 3:36.36 4:0.0
1 1:31.82 2:27.27 3:22.73 4:18.18
0 1:22.73 2:13.64 3:36.36 4:27.27
1 1:30.43 2:39.13 3:13.04 4:17.39 ......................
My problem is that when I count the number of lines in the output model generated by svm-train (svm_train_model.txt), this has 12 fewer lines than that of the input file. The line count here shows 450, although there are obviously also 9 lines at the beginning showing the various parameters generated
i.e.
svm_type c_svc
kernel_type rbf
gamma 1
nr_class 2
total_sv 441
rho -0.156449
label 0 1
nr_sv 228 213
SV
Therefore 12 lines in total from the original input of 453 have gone. I am new to svm and was hoping that someone could shed some light on why this might have happened?
Thanks in advance
Updated.........
I now believe that in generating the model, it has removed lines whereby the labels and all the parameters are exactly the same.
To explain............... My input is a set of miRNAs which have been classified as 1 and 0 depending on their involvement in a particular process or not (i.e 1=Yes & 0=No). The input file looks something like.......
0 1:22 2:30 3:14 4:16
1 1:26 2:15 3:17 4:25
0 1:22 2:30 3:14 4:16
Whereby, lines one and three are exactly the same and as a result will be removed from the output model. My question is then both why the output model would do this and how I can get around this (whilst using the same features)?
Whilst both SOME OF the labels and their corresponding feature values are identical within the input file, these are still different miRNAs.
NOTE: The Input file does not have a feature for miRNA name (and this would clearly show the differences in each line) however, in terms of the features used (i.e Nucleotide Percentage Content), some of the miRNAs do have exactly the same percentage content of A,U,G & C and as a result are viewed as duplicates and then removed from the output model as it obviously views them as duplicates even though they are not (hence there are less lines in the output model).
the format of the input file is:
Where:
Column 0 - label (i.e 1 or 0): 1=Yes & 0=No
Column 1 - Feature 1 = Percentage Content "A"
Column 2 - Feature 2 = Percentage Content "U"
Column 3 - Feature 3 = Percentage Content "G"
Column 4 - Feature 4 = Percentage Content "C"
The input file actually looks something like (See the very first two lines below), as they appear identical, however each line represents a different miRNA):
1 1:23 2:36 3:23 4:18
1 1:23 2:36 3:23 4:18
0 1:36 2:32 3:5 4:27
1 1:14 2:41 3:36 4:9
1 1:18 2:50 3:18 4:14
0 1:36 2:23 3:23 4:18
0 1:15 2:40 3:30 4:15
In terms of software, I am using libsvm-3.22 and python 2.7.5
Align your input file properly, is my first observation. The code for libsvm doesnt look for exactly 4 features. I identifies by the string literals you have provided separating the features from the labels. I suggest manually converting your input file to create the desired input argument.
Try the following code in python to run
Requirements - h5py, if your input is from matlab. (.mat file)
pip install h5py
import h5py
f = h5py.File('traininglabel.mat', 'r')# give label.mat file for training
variables = f.items()
labels = []
c = []
import numpy as np
for var in variables:
data = var[1]
lables = (data.value[0])
trainlabels= []
for i in lables:
trainlabels.append(str(i))
finaltrain = []
trainlabels = np.array(trainlabels)
for i in range(0,len(trainlabels)):
if trainlabels[i] == '0.0':
trainlabels[i] = '0'
if trainlabels[i] == '1.0':
trainlabels[i] = '1'
print trainlabels[i]
f = h5py.File('training_features.mat', 'r') #give features here
variables = f.items()
lables = []
file = open('traindata.txt', 'w+')
for var in variables:
data = var[1]
lables = data.value
for i in range(0,1000): #no of training samples in file features.mat
file.write(str(trainlabels[i]))
file.write(' ')
for j in range(0,49):
file.write(str(lables[j][i]))
file.write(' ')
file.write('\n')
I'm struggling with getting a simple correlation done. I've tried all that was suggested under similar questions.
Here are the relevant parts of the code, the various attempts I've made and their results.
import numpy as np
import pandas as pd
try01 = data[['ESA Index_close_px', 'CCMP Index_close_px' ]].corr(method='pearson')
print (try01)
Out:
Empty DataFrame
Columns: []
Index: []
try04 = data['ESA Index_close_px'][5:50].corr(data['CCMP Index_close_px'][5:50])
print (try04)
Out:
**AttributeError: 'float' object has no attribute 'sqrt'**
using numpy
try05 = np.corrcoef(data['ESA Index_close_px'],data['CCMP Index_close_px'])
print (try05)
Out:
AttributeError: 'float' object has no attribute 'sqrt'
converting the columns to lists
ESA_Index_close_px_list = list()
start_value = 1
end_value = len (data['ESA Index_close_px']) +1
for items in data['ESA Index_close_px']:
ESA_Index_close_px_list.append(items)
start_value = start_value+1
if start_value == end_value:
break
else:
continue
CCMP_Index_close_px_list = list()
start_value = 1
end_value = len (data['CCMP Index_close_px']) +1
for items in data['CCMP Index_close_px']:
CCMP_Index_close_px_list.append(items)
start_value = start_value+1
if start_value == end_value:
break
else:
continue
try06 = np.corrcoef(['ESA_Index_close_px_list','CCMP_Index_close_px_list'])
print (try06)
Out:
****TypeError: cannot perform reduce with flexible type****
Also tried .astype but not made any difference.
data['ESA Index_close_px'].astype(float)
data['CCMP Index_close_px'].astype(float)
Using Python 3.5, pandas 0.18.1 and numpy 1.11.1
Would really appreciate any suggestion.
**edit1:*
Data is coming from an excel spreadsheet
data = pd.read_excel('C:\\Users\\Ako\\Desktop\\ako_files\\for_corr_tool.xlsx') prior to the correlation attempts, there are only column renames and
data = data.drop(data.index[0])
to get rid of a line
regarding the types:
print (type (data['ESA Index_close_px']))
print (type (data['ESA Index_close_px'][1]))
Out:
**edit2*
parts of the data:
print (data['ESA Index_close_px'][1:10])
print (data['CCMP Index_close_px'][1:10])
Out:
2 2137
3 2138
4 2132
5 2123
6 2127
7 2126.25
8 2131.5
9 2134.5
10 2159
Name: ESA Index_close_px, dtype: object
2 5241.83
3 5246.41
4 5243.84
5 5199.82
6 5214.16
7 5213.33
8 5239.02
9 5246.79
10 5328.67
Name: CCMP Index_close_px, dtype: object
Well, I've encountered the same problem today.
try use .astype('float64') to help make the type correct.
data['ESA Index_close_px'][5:50].astype('float64').corr(data['CCMP Index_close_px'][5:50].astype('float64'))
This works well for me. Hope it can help you as well.
You can try as following:
Top15['Citable docs per capita']=(Top15['Citable docs per capita']*100000)
Top15['Citable docs per capita'].astype('int').corr(Top15['Energy Supply per Capita'].astype('int'))
It worked for me.
In my GUI I am using this matlab code to store the values in excel sheet.This code is calculating the glcm six features.
function [Contrast,cor,ener,homo,Var,Entropy] = glcm_feature_extraction(I1)
Contrast = graycoprops(graycomatrix(rgb2gray(I1)),'Contrast')
cor= graycoprops(graycomatrix(rgb2gray(I1)), 'Correlation')
ener = graycoprops(graycomatrix(rgb2gray(I1)), 'Energy')
homo = graycoprops(graycomatrix(rgb2gray(I1)), 'Homogeneity')
img = double(I1);
Var = var((img(:)))
Entropy=entropy(I1)
Here suppose I get these values after calculation:
[0.603606103 : 0.785092239 : 0.271728411 : 0.855429408 :1889.578963 : 7.178149206]
But iI need only approx value like:
[0.6 : 0.7 : .2 ....]
How to modify this code to get this result?
For example, lets consider Contrast=0.603606103
And you wanted to make approximately as 0.6 then use the following :
sprintf('%.1f',Contrast);
which should give you result exactly Contrast=0.6
Similarly do it for all remaining 5 variables.