I'm trying to code that calculates how many upper and lower characters in a string. Here's my code.
I've been trying to convert it to string, but not working.
def up_low(string):
result1 = 0
result2 = 0
for x in string:
if x == x.upper():
result1 + 1
elif x == x.lower():
result2 + 1
print('You have ' + str(result1) + ' upper characters and ' +
str(result2) + ' lower characters!')
up_low('Hello Mr. Rogers, how are you this fine Tuesday?')
I expect my outcome to calculate the upper and lower characters. Right now I'm getting "You have 0 upper characters and 0 lower characters!".
It's not adding up to result1 and result2.
Seems your error is in the assignation, missimg a '=' symbol (E.g. result1 += 1)
for x in string:
if x == x.upper():
result1 += 1
elif x == x.lower():
result2 +**=** 1
The problem is in the line result1 + 1 and result2 + 1. This is an expression, but not an assignment. In other words, you increment the counter, and then the incremented value goes nowhere.
The solution is to work the assignment operator = into there somewhere.
Related
Having an infinite sequence s = 1234567891011...
Let's find the number at the n position (n <= 10^18)
EX: n = 12 => 1; n = 15 => 2
import Foundation
func findNumber(n: Int) -> Character {
var i = 1
var z = ""
while i < n + 1 {
z.append(String(i))
i += 1
}
print(z)
return z[z.index(z.startIndex, offsetBy: n-1)]
}
print(findNumber(n: 12))
That's my code but when I find the number at 100.000th position, it returns an error, I thought I appended too many i to z string.
Can anyone help me, in swift language?
The problem we have here looks fairly straight forward. Take a list of all the number 1-infinity and concatenate them into a string. Then find the nth digit. Straight forward problem to understand. The issue that you are seeing though is that we do not have an infinite amount of memory nor time to be able to do this reasonably in a computer program. So we must find an alternative way around this that does not just add the numbers onto a string and then find the nth digit.
The first thing we can say is that we know what the entire list is. It will always be the same. So can we use any properties of this list to help us?
Let's call the input number n. This is the position of the digit that we want to find. Let's call the output digit d.
Well, first off, let's look at some examples.
We know all the single digit numbers are just in the same position as the number itself.
So, for n<10 ... d = n
What about for two digit numbers?
Well, we know that 10 starts at position 10. (Because there are 9 single digit numbers before it). 9 + 1 = 10
11 starts at position 12. Again, 9 single digits + one 2 digit number before it. 9 + 2 + 1 = 12
So how about, say... 25? Well that has 9 single digit numbers and 15 two digit numbers before it. So 25 starts at 9*1 + 15*2 + 1 = 40 (+ 1 as the sum gets us to the end of 24 not the start of 25).
So... 99 starts at? 9*1 + 89*2 + 1 = 188.
The we do the same for the three digit numbers...
100... 9*1 + 90*2 + 1 = 190
300... 9*1 + 90*2 + 199*3 + 1 = 787
1000...? 9*1 + 90*2 + 900*3 + 1 = 2890
OK... so now I'm seeing a pattern here that seems to need to know the number of digits in each number. Well... I can get the number of digits in a number by rounding up the log(base 10) of that number.
rounding up log base 10 of 5 = 1
rounding up log base 10 of 23 = 2
rounding up log base 10 of 99 = 2
rounding up log base 10 of 627 = 3
OK... so I think I need something like...
// in pseudo code
let lengthOfNumber = getLengthOfNumber(n)
var result = 0
for each i from 0 to lengthOfNumber - 1 {
result += 9 * 10^i * (i + 1) // this give 9*1 + 90*2 + 900*3 + ...
}
let remainder = n - 10^(lengthOfNumber - 1) // these have not been added in the loop above
result += remainder * lengthOfNumber
So, in the above pseudo code you can give it any number and it will return the position in the list that that number starts on.
This isn't the exact same as the problem you are trying to solve. And I don't want to solve it for you.
This is just a leg up on how I would go about solving it. Hopefully, this will give you some guidance on how you can take this further and solve the problem that you are trying to solve.
I understand the whole code and
I just want to know why there has to be a -1 at the end of the range function.
I've been checking it out with pythontutor but I can't make it out.
#Given 2 strings, a and b, return the number of the positions where they
#contain the same length 2 substring. So "xxcaazz" and "xxbaaz" yields 3,
#since the "xx", "aa", and "az" substrings appear in the same place in
#both strings.
def string_match(a, b):
shorter = min(len(a), len(b))
count = 0
for i in range(shorter -1): #<<<<<<<<< This is -1 I don't understand.
a_sub = a[i:i+2]
b_sub = b[i:i+2]
if a_sub == b_sub:
count = count + 1
return count
string_match('xxcaazz', 'xxbaaz')
string_match('abc', 'abc')
string_match('abc', 'axc')
I expect to understand why there has to be a -1 at the end of the range function. I will appreciate your help and explanation!
The value indices of the for loop are counted since 0 so the final value actually would be the (size -1)
5.2 Write a program that repeatedly prompts a user for integer numbers until the user enters 'done'. Once 'done' is entered, print out the largest and smallest of the numbers. If the user enters anything other than a valid number catch it with a try/except and put out an appropriate message and ignore the number. Enter 7, 2, bob, 10, and 4 and match the output below.
largest = None
smallest = None
while True:
num = input("Enter a number: ")
if num == "done":
break
print(num)
print("Maximum", largest)
and the output must be like this
Invalid input
Maximum is 10
Minimum is 2
somebody please help me with this????? enter image description here
I think this will do your work,
With simple if conditions
# Defining two variables to None.
largest = None
smallest = None
# starting an infinite loop
while True:
num = input("Enter a number: ")
# try block to capture ValueError
try:
if num == "done":
break
# assign the values of largest and smallest to num if its None ( on first iteration)
if largest is None:
largest = int(num)
if smallest is None:
smallest = int(num)
# changing the values of it greater or smaller
if int(num) > largest:
largest = int(num)
if int(num) < smallest:
smallest = int(num)
# capture the type error and ignores.
except ValueError:
print("ignored.")
continue
print("Maximum: " + str(largest))
print("Minimum: " + str(smallest))
By making use of list and it's methods
# Defining an empty list.
myList = []
# starting an infinite loop
while True:
num = input("Enter a number: ")
# try block to capture ValueError
try:
if num == "done":
break
# append the entered number to list if valid
myList.append(int(num))
# catches value error and ignores it
except ValueError:
print("ignored.")
continue
# prints maximum and min
if len(myList) > 0:
print("Maximum: " + str(max(myList)))
print("Minimum: " + str(min(myList)))
else:
print("List is empty")
largest = None
smallest = None
while True:
try:
num = input("Enter a number: ")
if num == "done":
break
num = int(num)
if largest is None or largest < num:
largest = num
elif smallest is None or smallest > num:
smallest = num
except ValueError:
print("Invalid input")
print ("Maximum is", largest)
print ("Minimum is", smallest)
largest = None
smallest = None
while True:
num = input("Enter a number: ")
if num == "done" :
break
try:
n = int(num)
except:
print ('Invalid input')
continue
if largest is None or n > largest:
largest = n
if smallest is None or n < smallest:
smallest = n
print("Maximum is", largest)
print("Minimum is", smallest)
#Print out largest and smallest number
largest=None
smallest=None
while True:
n = input('Enter a number: ')
if n == "done":
break
try:
num=int(n)
if largest is None:
largest = num
elif num > largest:
largest = num
if smallest is None:
smallest = num
elif num < smallest:
smallest = num
except:
print("Invalid input")
print('maximum:', largest)
print('Minimum:', smallest)
i got the error of this code which is:
path[index][4] += 1
IndexError: list index out of range
why this happened?how can i remove this error ?
Code:
def stress_centrality(g):
stress = defaultdict(int)
for a in nx.nodes_iter(g):
for b in nx.nodes_iter(g):
if a==b:
continue
pred = nx.predecessor(G,b) # for unweighted graphs
#pred, distance = nx.dijkstra_predecessor_and_distance(g,b) # for weighted graphs
if a not in pred:
return []
path = [[a,0]]
path_length = 1
index = 0
while index >= 0:
n,i = path[index]
if n == b:
for vertex in list(map(lambda x:x[0], path[:index+1]))[1:-1]:
stress[vertex] += 1
if len(pred[n]) >i:
index += 1
if index == path_length:
path.append([pred[n][i],0])
path_length += 1
else:
path[index] = [pred[n][i],0]
else:
index -= 1
if index >= 0:
path[index][4] += 1
return stress
Without the data it's hard to give you anything more than an indicative answer.
This line
path[index][4] += 1
assumes there are 5 elements in path[index] but there are fewer than that. It seems to me that your code only assigns or appends to path lists of length 2. As in
path = [[a,0]]
path.append([pred[n][i],0])
path[index] = [pred[n][i],0]
So it's hard to see how accessing the 5th element of one of those lists could ever be correct.
This is a complete guess, but I think you might have meant
path[index][1] += 4
I have converted a decimal number to binary using STR$() in QBASIC. But I need a way to convert decimal number to binary without using string functions. Thanks.
My Code :
CLS
INPUT N
WHILE N <> 0
E = N MOD 2
B$ = STR$(E)
N = FIX(N / 2)
C$ = B$ + C$
WEND
PRINT "Output "; C$
END
This code sample converts a numeric value to a binary string in Basic.
PRINT "Enter value";
INPUT Temp#
Out3$ = ""
IF Temp# >= False THEN
Digits = False
DO
IF 2 ^ (Digits + 1) > Temp# THEN
EXIT DO
END IF
Digits = Digits + 1
LOOP
FOR Power = Digits TO 0 STEP -1
IF Temp# - 2 ^ Power >= False THEN
Temp# = Temp# - 2 ^ Power
Out3$ = Out3$ + "1"
ELSE
Out3$ = Out3$ + "0"
END IF
NEXT
END IF
PRINT Out3$
END
When you want to display an integer value as binary, it seems logical to me to store it in a string variable, because it's only for display. So I'm not really sure what you are trying to do here.
Maybe you were looking for LTRIM$ so you would get outputs like 11010 instead of 1 1 0 1 0 ?
You could store it in an integer value like in the code below. But, although the integer value will look the same as the string variable, it will in fact be a completely different value.
CLS
INPUT "Type a decimal number:", N
S$ = ""
I = 0
P = 1
WHILE (N <> 0)
' get right most bit and shift right
E = N AND 1
N = INT(N / 2) ' bit shift right
' format for dsplay
S$ = LTRIM$(STR$(E)) + S$
I = I + (E * P)
P = P * 10
WEND
PRINT "Binary as string="; S$
PRINT "Binary as int="; I
END