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oplossingen extra oefeningen programmeren python dawynt dodona H5: functions
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Oplossingen extra gemaakte oefeningen in dodona op reeks 5 (functies). Programmeren van peter dawnyt ugent.
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H5: FUNCTIESKopzorgen voor Noach
def splitsing(woord):
klinkers = 'aeiouAEIOU'
pre x = ""
i=0
while woord[i] not in klinkers:
pre x += woord[i] #pre x van medeklinkers
i += 1
lengte_pre x = len(pre x)
su x = woord[lengte_pre x:]
return (pre x, su x)
def kruising(woord1,woord2):
pre x1, su x1 = splitsing(woord1)
pre x2, su x2 = splitsing(woord2)
kruising1 = pre x1 + su x2
kruising2 = pre x2 + su x1
return (kruising1, kruising2)
McArthurs wiskundetruc
def gast(maand,leeftijd):
maanden = ["januari",
"februari","maart","april","mei","juni","juli","augustus","september","november","december"]
rangnr = maand
getal = ((((rangnr * 2) + 5) * 50) + int(leeftijd)) - 365
return getal
def macArthur(getal):
nieuw = getal + 115
string = str(nieuw)
lengte = len(string)
if len(string) == 4:
rangnr = string[0] + string[1]
leeftijd = string[2:]
else:
rangnr = string[0]
leeftijd = string[1:]
return (int(rangnr), int(leeftijd))
Levensverwachting
def levensverwachting(geslacht,roker,sport,alcohol,fastfood):
som = 70
if geslacht == "vrouw":
som += 4
if roker is True:
som -=5
else:
som += 5
if sport == 0:
som -=3
else:
som += (1) * int(sport)
if alcohol ==0:
ffifi fi fiffi fi ffi fiffifi
, som +=2
elif alcohol <=7:
som +=0
else:
som -= 0.5 * (int(alcohol) - 7)
if fastfood is False:
som += 3
return oat(som)
Woordsommen
def letterwaarde(letter):
if letter.isalpha():
klein = letter.lower()
waarde = ord(klein) - ord('a') + 1 #positie in alfabet
else:
waarde = 0
return waarde
def woordwaarde(woord):
som = 0
for karakter in woord:
waarde = letterwaarde(karakter)
som += waarde
return som
def iswoordsom(woord1,woord2,woord3):
return (woordwaarde(woord1) + woordwaarde(woord2) == woordwaarde(woord3))
# zal true of false returnen
123
def evenOneven(n):
even = 0
oneven = 0
for karakter in str(n):
if int(karakter) % 2==0:
even +=1
else:
oneven +=1
return (even, oneven)
#procedure eenmaal toepassen
def volgende(n):
even, oneven = evenOneven(n)
return int(str(even) + str(oneven) + str(even + oneven))
def stappen(n):
stappen = 0
while n !=123:
n = volgende(n)
stappen += 1
return stappen
Meirp
def omgekeerd_getal(getal):
return int(str(getal)[::-1])
fl
, def ispriem(getal):
for i in range(2,int(getal)):
if int(getal) % i ==0:
return False
return True
def ismeirp(getal):
omgekeerd = omgekeerd_getal(getal)
priem = ispriem(getal)
priem2 = ispriem(omgekeerd)
if (priem is True) and (priem2 is True) and (str(omgekeerd) != str(getal)):
return True
else:
return False
Bovensteboven
def bovensteboven(getal):
ok = "01689"
nietok = "2357"
woord = ""
for karakter in str(getal):
if karakter in ok:
if karakter == "6":
woord += "9"
elif karakter == "9":
woord += "6"
else:
woord += karakter
omg = woord[::-1]
if omg == str(getal):
return True
else:
return False
def volgende(n):
getal = n + 1
waarde= bovensteboven(getal)
while waarde is False:
getal += 1
waarde = bovensteboven(getal)
return getal
C-som
def csom(getal):
if len(str(getal)) ==1:
return getal
else:
while len(str(getal)) > 1:
som = 0
for karakter in str(getal):
som += int(karakter)
getal = som
return getal
Hardnekkigheid
def vermenigvuldiging(getal):
som = 1
for karakter in str(getal):
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