fbrundu

import fastcluster as fc
import pandas as pd
import scipy.cluster.hierarchy as sch

# define total number of cluster to obtain
k = 5

# define matrix path
mat_path = 'matrix.txt'

fbrundu

import pandas as pd

# load data
mat = pd.read_table('class_matrix.txt', index_col=0)

# initialize consensus array
consensus_a = pd.Series(index=mat.index)

# define columns subset on which compute consensus
# in this case all columns are used

fbrundu

import numpy as np
import pandas as pd

# load data
mat = pd.read_table('matrix.txt', index_col=0)

# get classes
classes = np.unique(mat.values)
classes = classes[~np.isnan(classes)]

fbrundu

" no vi-compatible
set nocompatible

let g:python_host_prog=$HOME.'/.pyenv/versions/neovim2/bin/python'
let g:python3_host_prog=$HOME.'/.pyenv/versions/neovim3/bin/python'

" Setting up Vundle - the vim plugin bundler
let iCanHazVundle=1
let vundle_readme=expand('~/.vim/bundle/vundle/README.md')
if !filereadable(vundle_readme)

fbrundu

# line number after ==
# change \\t with the delimiter you want to use instead of tab for splitting to words
words_count=$(awk 'NR==5{print}' data.csv | tr \\t \\n | wc -l)
unique_words_count=$(awk 'NR==5{print}' data.csv | tr \\t \\n | uniq | wc -l)

if [[ "$words_count" -ne "$unique_words_counts" ]]; then
    echo "Not unique"
else
    echo "Unique"

fbrundu

A poker hand consists of 5 cards dealt at random without replacement from a standard deck of 52 cards of which 26 are red and the rest black. A poker hand is dealt. Find the chance that the hand contains three red cards and two black cards.

To achieve it, we use the [hypergeometric][1] probability mass function. We want 3 cards from the set of 26 red cards and 2 from the set of 26. So the parameters for the hypergeometric function are:

M = 52  # Total number of cards
n = 26  # Number of Type I cards (e.g. red cards) 
N = 5   # Number of draws (5 cards dealt in one poker hand)
k = 3   # Number of Type I cards we want in one hand

fbrundu

If you bet on "red" at roulette, you have chance 18/38 of winning. Suppose you make a sequence of independent bets on “red” at roulette, with the decision that you will stop playing once you have won 5 times. What is the chance that after 15 bets you are still playing?

We use [binomial][1] probability mass function. To calculate the probability, you have to estimate the probability of having up to 4 successful bets after the 15th. So the final probability will be the sum of the probability to get 0 successful bets in 15 bets, plus the probability to get 1 successful bet, ..., to the probability of having 4 successful bets in 15 bets.

To achieve it:

import scipy.stats as ss

n = 15         # Number of total bets

p = 18./38 # Probability of getting "red" at the roulette

fbrundu

import pandas as pd
import sys
import glob
import os

# input / output directory
input_dir = sys.argv[1]
# input file extension
input_ext = sys.argv[2]
# cardinality of index columns (rownames)

fbrundu

import pandas as pd
import numpy as np
import sys
import random as rnd

csv = sys.argv[1]
out = sys.argv[2]

df = pd.read_table(csv, sep='\t', index_col=0)

fbrundu

import pandas as pd
import sys
import re

tcga_tsv = sys.argv[1]

tcga = pd.read_table(tcga_tsv, sep='\t', index_col=0)

oldcolumns = tcga.columns.tolist()
newcolumns = ['-'.join(re.findall(r'TCGA[^_]*', oc)[0].split('-')[:4])