jmoy

counters = {key:collections.Counter() for key in ['child','mother','father']}
with open("day2/names.csv") as f:
    rdr = csv.DictReader(f)
    for r in rdr:
        for key in counters:
            l = r[key].split()
            if len(l)>0:
                counters[key][l[0]] += 1
                
commonize = lambda ctr,N: {t[0] for t in ctr.most_common(N)}

jmoy

#include <cstdio>
#include <cstdlib>
#include <vector>
#include <unordered_map>
#include <utility>
using namespace std;

#include <time.h>
#include <sys/time.h>
#include <sys/resource.h>

jmoy

memlimited: memlimited.cc calc_mean.cc calc_mean.h
	g++ -o memlimited -O3 -march=native -fopenmp memlimited.cc calc_mean.cc

jmoy

memlimited: memlimited.cc calc_mean.cc calc_mean.h
	clang++ -o memlimited -O3 --std=c++17 \
            -ffast-math -march=native \
            memlimited.cc calc_mean.cc -pthread

jmoy

section ‹A Simple Graph Problem›

text ‹
We shall prove the following: "In a finite group of people, some of whom are friends with some
of the others there must be at least two people who have the same number of friends."
›

theory Friends
  imports   Main 
            Finite_Set 

jmoy

clear all
frame create maternal_depr
frame maternal_depr {
	use "IABR7EFL.DTA"
	keep if midx==1
	gen no_trained_assist = (m3a==0 & m3b==0 & m3c==0)
	gen few_antenatal = m14<4
	gen missed = missing(m14) | missing(m3a) | missing(m3b) | missing(m3c) | m14>=98 
	collapse (max) no_trained_assist few_antenatal missed, by(v001 v002)
	gen deprived = no_trained_assist|few_antenatal