This program lets the user enter a string and displays the character that appears most frequently in a string.
I need help explaining frequent = i.
# This program displays the character that appears most frequently in the string
def main():
# Local variables.
count = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
index = 0
frequent = 0
# Get input.
user_string = input('Enter a string: ')
for ch in user_string:
ch = ch.upper()
# Determine which letter this character is.
index = letters.find(ch)
if index >= 0:
# Increase counting array for this letter.
count[index] = count[index] + 1
# Please help me explain this entire part!
for i in range(len(count)):
if count[i] > count[frequent]:
frequent = i
print('The character that appears most frequently' \
' in the string is ', letters[frequent], '.', \
sep='')
# Call main
main()
The code snippet in question:
for i in range(len(count)):
if count[i] > count[frequent]:
frequent = i
First the for loop iterates over the length of count which is 26.
The if statement:
if count[i] > count[frequent]:
Checks to see if the current letter in the for loop is larger than the current most frequent character. If it is then it sets the new most frequent character as the index of the for loop.
For example,
If A is referenced 12 times and B is referenced 14 then on the second loop when i = 1 the if statement would look like this:
if 12 > 14:
frequent = 1
This sets frequent to 1 which can be used to find the frequency in count for ex.
count[1] == 14
There are 26 different items in the list count, and 26 letters in the charset. It iterates through the count list for each item (that's the for i in range (len(count)) part) and then sees if the value of that item is greater than the value of the current largest item it's found - simply speaking it finds the largest value in the array, but instead of getting the value it gets the index, frequent = i is setting the index of the largest value currently found as it iterates to the variable frequent. It's simpler and more pythonistic to simply do
frequent = index(max(count)
which has EXACTLY the same effect
Related
Here I have a string in a list:
['aaaaaaappppppprrrrrriiiiiilll']
I want to get the word 'april' in the list, but not just one of them, instead how many times the word 'april' actually occurs the string.
The output should be something like:
['aprilaprilapril']
Because the word 'april' occurred three times in that string.
Well the word actually didn't occurred three times, all the characters did. So I want to order these characters to 'april' for how many times did they appeared in the string.
My idea is basically to extract words from some random strings, but not just extracting the word, instead to extract all of the word that appears in the string. Each word should be extracted and the word (characters) should be ordered the way I wanted to.
But here I have some annoying conditions; you can't delete all the elements in the list and then just replace them with the word 'april'(you can't replace the whole string with the word 'april'); you can only extract 'april' from the string, not replacing them. You can't also delete the list with the string. Just think of all the string there being very important data, we just want some data, but these data must be ordered, and we need to delete all other data that doesn't match our "data chain" (the word 'april'). But once you delete the whole string you will lose all the important data. You don't know how to make another one of these "data chains", so we can't just put the word 'april' back in the list.
If anyone know how to solve my weird problem, please help me out, I am a beginner python programmer. Thank you!
One way is to use itertools.groupby which will group the characters individually and unpack and iterate them using zip which will iterate n times given n is the number of characters in the smallest group (i.e. the group having lowest number of characters)
from itertools import groupby
'aaaaaaappppppprrrrrriiiiiilll'
result = ''
for each in zip(*[list(g) for k, g in groupby('aaaaaaappppppprrrrrriiiiiilll')]):
result += ''.join(each)
# result = 'aprilaprilapril'
Another possible solution is to create a custom counter that will count each unique sequence of characters (Please be noted that this method will work only for Python 3.6+, for lower version of Python, order of dictionaries is not guaranteed):
def getCounts(strng):
if not strng:
return [], 0
counts = {}
current = strng[0]
for c in strng:
if c in counts.keys():
if current==c:
counts[c] += 1
else:
current = c
counts[c] = 1
return counts.keys(), min(counts.values())
result = ''
counts=getCounts('aaaaaaappppppprrrrrriiiiiilll')
for i in range(counts[1]):
result += ''.join(counts[0])
# result = 'aprilaprilapril'
How about using regex?
import re
word = 'april'
text = 'aaaaaaappppppprrrrrriiiiiilll'
regex = "".join(f"({c}+)" for c in word)
match = re.match(regex, text)
if match:
# Find the lowest amount of character repeats
lowest_amount = min(len(g) for g in match.groups())
print(word * lowest_amount)
else:
print("no match")
Outputs:
aprilaprilapril
Works like a charm
Here is a more native approach, with plain iteration.
It has a time complexity of O(n).
It uses an outer loop to iterate over the character in the search key, then an inner while loop that consumes all occurrences of that character in the search string while maintaining a counter. Once all consecutive occurrences of the current letter have been consumes, it updates a the minLetterCount to be the minimum of its previous value or this new count. Once we have iterated over all letters in the key, we return this accumulated minimum.
def countCompleteSequenceOccurences(searchString, key):
left = 0
minLetterCount = 0
letterCount = 0
for i, searchChar in enumerate(key):
while left < len(searchString) and searchString[left] == searchChar:
letterCount += 1
left += 1
minLetterCount = letterCount if i == 0 else min(minLetterCount, letterCount)
letterCount = 0
return minLetterCount
Testing:
testCasesToOracles = {
"aaaaaaappppppprrrrrriiiiiilll": 3,
"ppppppprrrrrriiiiiilll": 0,
"aaaaaaappppppprrrrrriiiiii": 0,
"aaaaaaapppppppzzzrrrrrriiiiiilll": 0,
"pppppppaaaaaaarrrrrriiiiiilll": 0,
"zaaaaaaappppppprrrrrriiiiiilll": 3,
"zzzaaaaaaappppppprrrrrriiiiiilll": 3,
"aaaaaaappppppprrrrrriiiiiilllzzz": 3,
"zzzaaaaaaappppppprrrrrriiiiiilllzzz": 3,
}
key = "april"
for case, oracle in testCasesToOracles.items():
result = countCompleteSequenceOccurences(case, key)
assert result == oracle
Usage:
key = "april"
result = countCompleteSequenceOccurences("aaaaaaappppppprrrrrriiiiiilll", key)
print(result * key)
Output:
aprilaprilapril
A word will only occur as many times as the minimum letter recurrence. To account for the possibility of having repeated letters in the word (for example, appril, you need to factor this count out. Here is one way of doing this using collections.Counter:
from collections import Counter
def count_recurrence(kernel, string):
# we need to count both strings
kernel_counter = Counter(kernel)
string_counter = Counter(string)
# now get effective count by dividing the occurence in string by occurrence
# in kernel
effective_counter = {
k: int(string_counter.get(k, 0)/v)
for k, v in kernel_counter.items()
}
# min occurence of kernel is min of effective counter
min_recurring_count = min(effective_counter.values())
return kernel * min_recurring_count
so i am doing the cs50 dna problem and i am having a hard time with the counter as i don't know how to code to properly count the highest amount of times the sequence i am looking for has repeated without another sequence in between the two. for example i am looking for the sequence AAT and the text is AATDHDHDTKSDHAATAAT so the highest amount should be two as the last two sequence is AAT and there is no sequence between them.
here is my code:
text="TCTAGTCTAGTCTAGTCTAGTCTAGACTTGTCGCTGACTCCGAGAAGATCCTAACATTAACCAATTCCCCCTAGTCTGAGGCACGGTTACCGATCGGGTTAATGGATCTCTCACCGTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAAACGTGTAACTGTAATAATCCGCCCGAAAAAACTGATCTTAGGGTTGCGGCATCTGCACGTGACAGTGTGCTACTGTTAGATAGAGGGATCAAACGAGGTTGCAAGGATTATATCTCTCCGTGCTCGATAAGACACAGCCGGTTGCGGGCTGCTTCCTCTGGATCCAATGCAGCCGTACGTACACCGTAGAGCAAATTTAGTGGTAAAGGAACTTGCTCAAACACTACGGCTTCGGGCTACTGTTGGCGCCGGTTGGGGATCCCATTCAACGCTGGCCCTTTCGCTATGGTTCGGTGATTTTACACCGAAGCGAACCTTGAACCGTGGATTTCGGGTGTCCTCCGTTTTTAGGTACTGCGTGCAGACATGGGCACCTGCCATAGTGCGATCAGCCAGAATCCATTGTATGGGAGTTGGACTCGTTTGAATTTACCGGAAACCTCATGCTTGGTCTGTAGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGAAACTGGGCGACTTGAAGTCGGCTTGCGTATTAATAGCTCTGCAATGTAACTCGGCCCTTGGCGGCGGGCAGCTTAGTATTGAACCGCGACACACCATAGGTGCGGCAAATATTAAAAGTACGCTCGAACCGGAACCTGTCTCCATGACTGGACGACCAGCCCGGCGTCTTCTACGTAACACAGGGGGCTGTCGAGGTAGGGCGTAGGAACTTCGGGGTCACTACGCCGTAACAGCACCGAATATCATATCATCCAACTTGCTTGGTACATGCCCCGTTCTGTATCAAAAGTTTACGGCCCCGGACATACCTGCTGTCAGTTGAATACCTATGCGAGTCTGAAACACGAATAGTTCAGGCGTGCAAAGACACGCTAAGCACACGCCGCAGGCAGGGGGGGTATTTTATAAGTCGTTTTTTGGAAGGGTAATGTAAACTTATCCCATAATACCCTTTGGCTTCCCCTCACTCGTGCACTTCTCATAATGATACGTCAGGGTGATTGTAGATTCACGCGTCATCAGATTGTCCCTTTCTCGAGTCTTAGTATCTTTCCTAATCCGCTCGACTCTGCGCCATGATCGAATTCCTGACAGGCTACAAGAATAAACTGCCAGCATACTCCTTACCGATTGGCGCCTACTAATTATACGCACATGGGCATCTTCGACGTCTAAACATAGGCTCTTAGTATTCCGTAGGATGTTGAGCCGACAGGAAAGTCAAACGTCGTGGGTGACCGTAGCCTGACTCGCCCGACGCAGGATTCGCTCATATGTGTGAACGGATGCTTATGTAACTTCCTAATTGCAGCGAATGGCAGTTCCGTAGTGAAGGTTCGAAACGTACGGGGTCCGGCCATGGATTAGATCTTTCAGTGCGCTAAACTCTTAACCGCAGATACTTGGCGGACCATCTTCGTGTTGCTACTATGGTATAGACCAGGCTGTCGAATCTACTTAACACAGGTGAACCCCCAGATCGGCTAGAGCCTTCGAGGCTAGACCTTTAACAATCTTTAGACACTTCCAAATCGCGGCCGGATATGTCTCGTTGGCAGCCGCAGACAAGAGAAGAGGGTCGGCAGTGTCTGCCACGCGTGACCTGTATGATCTTAGCCTTTAAGATCACACTACTGATCACAATCTATTATGATTGCCTTAGCTAACTGAGTGATGCACCCCCACAGGCTGAGAGAAATCTGTAGTTTGACGACACGCCGTCTGGCTAAAAATGTGAATCCGCCGATCCGAGACGGTGGAAGCTTGAGACCAAATGCGGGAAACCAATGACTTCATTACGGAACAAGACATAACGGCGTGAGTTGACGACTGGGATTAACCCTTTCCCGAGTCTGTACTTCTGCTACACAATGAGGATGCGAATTATCTAAGACCTTGTACTACCTAAACTAACCCTGAGGCGGGCATTGAATTCCGGCCATCTTCAGCCCAAAGAAAGACCAAATGTGAGGAAAATGAGGGATCGGTATAAGCTTTTCACGATCTCAAGGTTCACGGCCGCCAGGGCCGTAGTTGGGGCTTCATGCACATTGCCAACCCGGACATCGACAGTCGGTACCGCAGGGGTTCGAGGAATACTCCCAGCTGTGACACCTGGTCGTCGACTGGACCCAGCTGGTGGGCGGCATAGGTAGTTAATACTGAATTAAAGCCGGGAACGTCTCTCTAACTAGAAACCTTGTGATAGGATACACAGACCTAGTGCCCCGACGTTAGCATTTGAATTCATCTATCTTGGCGTCTTTTAGTAGGCCTGGGTCAACTCCGGCGTTGGCCAAAATAACCGATCTGCGTTATGTGGCCACGCATCGAGTGACAGGGTGCATACAAATTGATGGTCAAAGAGTTTAAACAAGACAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCCCCACGCTTCTACATAGCCACACTGGAGCTAGTCCTCGTGTTAAATTTTTCGCTTGTTGCACGGTTATCATCAGAAGTGCCACTGGTATTCCTCTGTAGCTCCCGTATGCCGAAGGTTGCGGCTTAGGTACTGCTTATACACGTCTCTCAAGTTTGTCAGCCGCGTGATCTTTCTGCGGGGATAGGTGATCGTCCCTCGCTCCGGACATTGCATTAAAATTACCTAGTTGATAGGGCGGCGGAGTTGCATACCGGCGTTCAATCGCGGCTCCAGACTGGTTTGAGCTACGCGTCTGCCAGCGTGAAAAAGCTGATTTGTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCCAGGTATTATCATTTGAATCGTATGTTTTCTGCCGTACGTCAACTGCGTCGTCGGGGACTGAAATGGTCTGCCTCCAGACCCTTACCTCCCGATAAGCCATGACTAAGTATGTGAAGGATCACCTGAATTGCTGAAAGTTAACGGTAAGATATCTGAAAGAGCTCATTAGATCCAACACTTATCTACTCAAAAATTCGTCATATTTCGGTGACTTGCTAGAAAGGCTCTTGCACAGTAAGGTTATAGAGAATGCTACCGTTGAAGCACCAGCCGTTGAAGCCCGCCTTTAACCACGCGATATATCCAATTAACCAAGGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGTCGCCTTGTAATAATTACTTTGGCCCGGATTATAACGAAGGAACTCGCCATGAACTCGCAGCACGTTGTACTGGAACAATCTACTTTTTATAATATAGCGATAACTCCCAGCTTTTATGTGGGTGATATTGTCCTAGCTTTTTAAAGATACCCTCTGGCCCGGTCCAAGTAAGGTCCACATTGCCTGACGTAAGCGTACGGTCAACGGGTGCACCGGTTCCCGCTAAAGCTCGATCCTATTCTTTCAGTCGGGGGGAAATAAACTCGTATACTCTCCACCCACCCGTACGTCCCGGACTAGAATAACTACCGGGTATTTCCGGTTCGTAACACCACGCCATGACGTGTCAACATAAACGCTTCTTTTGAAAGGTGCACATGCAGATTGCACAAGCAGCAGGCACCGCCCTTATCCATATCCTGTTGAGGCCCTCGATCCTAGTGTTCCTTGTTATCAGGATATTTTCTCGCTGTACGTTATTGTCCTTTTCAAATTACAACTGACCGCTTCCTCACCCGCTAAACCCTACCTTACGCACAACCAAGGCCTTGTCCCGGATGAACCCGGCTGCTCCTATGGATAAGCAACCCAGCCCGGCAGTTTACTTCAGGTGTTATCGTCGACTGACACCCTCAGCTTTCTCCCATTACACAGCGAGTATTTTCCGCGTAGCAATGGCAGTGACTTTGAGCGCACACTCAGAAGCCGTTGGAATGGCACCGGGGACGGCCCGATTTAGCCCCGCACACCTCCTGGAATCTTAGATCGCACGGCGATCTCGGTTCAGGCACCAACCCCAAAGAGTGTTTTGAGTTTTTGGTATGGCTCGCCTCAATTATCGGTTTTCGCTGCTCTGTGCCTGTCAACTCGGCTAGCTGTCGTGTTTTGTCGATCAGTGCGTGGACACTCTCGGTCGATGGTCGTGGATGGGACTGTAGTAAGTTTCACCGAAGCAGGAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAACTTCGCTTCATATAACGTAGCCATAGTGCTGTCTGCCATCAATAAGTCTTGCTCAGTGGTGCATACGTCGGGGAGGTTTGTTCCGCCTGGTCAGAACGAGTCTAGGGCGAGCCTATAGGCCAGTCGAGAGCCAAGATTCTATGAAATTAATACGACTACTGGGTGAGAGGTCATACAATTCCCGTGGAATCTGTACCTAAGATATTTCCAGATAGGGATGGCTACTGGTTAAGTTGACAGTGTCTAGATACGTGAGAGCACCTGAGAGGACGCCACGAGTCGGAGCGTGGGCGATCACCCTTCTGAGTCATAAGTCATGTCTATATATCCCTCACTAAAAAGGGCACACGACTATACATGCTTGAGCTTTACGGTCTGGCATGTGGAATGCCCGGAGCAACCCAGTCTTACCATCCTTTACGTACATTTACCGACCCGGCAGTGGCCGGCGCGGAAACCCAGGAGAACGTCGGTCATGATACGCGCCCTCCGCCGAAAGCGTGCTCACACCTCAGGATATCAGCGCTATTACCGGACGTCCCGCGTCCACCATCTAATAATTCAGGTGCTCCTAATAAGTGGGCTGGAGAGCGAGGATTGATATACGTTGAGGAGCTCCGACGGCCCTCTCGTGCGTTTGATGTAGATTGCGTTACCGACGGAGCACGCGTTTGTCAATTTCTGTCTAGGGACGTTTATGTCCTCAATACGAATACCAGGCCTATTTTAGTGTACAAATCACTTAGCAGTCGGAATTGGAAACCTGATGGAAGCGT"
counter=0
length=len(text)
search="AGATC"
tmp=0
for i in range(length):
if text[i:i + len(search)] == search:
tmp += 1
if tmp > counter:
counter = tmp
if text[i:i + len(search)] != search:
tmp = 0
print("done")
print(counter)
try this
import re
sequence = "AATDHDHDTKSDHAATAAT"
matches = re.findall(r'(?:AAT)+', sequence)
largest = max(matches, key=len)
print(len(largest)//len('AAT'))
basically this way will find you the list of substrings in the string you are have then you choose the largest substring. The number of occurrences of the substring will be the length of the largest divide by the length of substring
First and foremost, the regex solution is the Python way to solve this. However, if you want your code repaired ...
The problem with your code is that your index fails to acknowledge that you've found a match. You have no way to recognize consecutive occurrences.
Consider the case where you've found the start of a triple-match, AATAATAAT. You get to the first A, recognize, the AAT and increment tmp. You go to the next loop iteration, and now i points at the second A. You see that it's not AAT here (it's ATA, spanning the first two occurrences), so you record one instance and reset all your status variables.
Instead, you have to jump to the end of the first match and look for a second. Since your index does not move smoothly by increments of 1, you'll want a while loop instead.
Please learn to use meaningful variable names where the variables have
any meaning. i is fine if all it does is to manage your loop. As
soon as you use it for anything else, give it a real name. Similarly,
tmp and count really need replacing.
snip_size = len(search)
pos = 0 # position in the genetic sequence
rep = 0 # number of consecutive repetitions
max_rep = 0 # longest repetition sequence found
while pos < length:
if text[pos:pos + snip_size] == search:
rep += 1
pos += snip_size
else:
max_rep = max(max_rep, rep)
rep = 0
pos += 1
print(max_rep, "repetitions found")
Output:
15 repetitions found
I am trying to write a python program that will take any string of lowercase letters and return the longest alphabetical substring within it. Below is a segment of the code.
s="abc" #sample string
anslist=[] #stores answers
shift=0 #shifts substring
expan=0 #expands substring
while len(s) >= 1+shift+expan: #within bounds of s
if s[0+shift+expan] > s[1+shift+expan]: #if not alphabetical
shift += 1 #moves substring over
else: #if alphabetical
while s[0+shift+expan] <= s[1+shift+expan]: #while alphabetical
expan += 1 #checks next letter
anslist += s[0+shift:2+shift+expan] #adds substring to ans
expan = 0 #resets expansion
When I run the code, the line containing
while s[0+shift+expan] <= s[1+shift+expan]:
creates an error that the string index is outside of the range. I see that adding to expan will put the index out of range, but shouldn't the largest while loop solve this? I appreciate any help.
First, why your code doesn't work:
You aren't protecting your inner loop against running off the end of the string
your indexes are off when "saving" the substring
you += onto anslist, which is not how you add strings to a list
you don't increment the shift after processing a substring, so when it clears expan it starts over at the same index and loops forever
Fixed code (inline comments explain changes):
s="abckdefghacbde" #sample string
anslist=[] #stores answers
shift=0 #shifts substring
expan=0 #expands substring
while len(s) > 1+shift+expan: #within bounds of s
if s[0+shift+expan] > s[1+shift+expan]: #if not alphabetical
shift += 1 #moves substring over
else: #if alphabetical
# Added guard for end of string
while len(s) > 1 + shift + expan and # While still valid
s[0+shift+expan] <= s[1+shift+expan]:# While alphabetical
expan += 1 #checks next letter
# Fixed string sublength and append instead of +=
anslist.append(s[0+shift:1+shift+expan]) #adds substring to ans
# Continue to next possible substring
shift += expan # skip inner substrings
expan = 0
print anslist
Results in:
['abck', 'defgh', 'ac', 'bde']
So the last step would be to find the one with the longest length, which I'll leave up to you, since this looks like homework.
To answer the question:
I see that adding to expan will put the index out of range, but shouldn't the largest while loop solve this?
It protects against your starting substring index from going off, but not your expansion. You must protect against both possibilities.
Have a look at this.
>>> import re
>>> words = []
>>> word = r'[a]*[b]*[c]*[d]*[e]*[f]*[g]*[h]*[i]*[j]*[k]*[l]*[m]*[n]*[o]*[q]*[r]*[s]*[t]*[u]*[v]*[x]*[y]*[z]*' # regex that would match sub-strings. Just extend it up to z.
>>> string = "bacde"
>>> for m in re.finditer(word, string):
... words.append(m.group())
>>> print(words) # list of substrings
['b', 'acde']
Then you can extract the largest string from the list of strings
>>> print(max(words, key=len))
acde
For an assignment, I need to use a while loop to reverse a list, and I just can't do it.
This is the sample code I have to help me get started:
sentence = raw_int (" ")
length = len(sentence) # determines the length of the sentence (how many characters there are)
index = length - 1 #subtracts one from the length because we will be using indexes which start at zero rather than 1 like len
while... #while the index is greater than or equal to zero continue the loop
letter = sentence[index] #take the number from the index in the sentence and assigns it to the variable letter
I need to use this in my solution.
sentence = raw_input(" ")
length = len(sentence)
index = length - 1
reversed_sentence = ''
while index >= 0:
#letter is the last letter of the original sentence
letter = sentence[index]
#make the first letter of the new sentence the last letter of the old sentence
reversed_sentence += letter
#update the index so it now points to the second to last letter of the original sentence
index = index - 1
print reversed_sentence
Because this is an assignment, I'm not going to give you the full code. But I will give you two 'hints'.
1) a sentenced is reversed if every character is 'flipped'. For example, 'I ran fast'-to flip this sentence first swap 'I' and 'f', then space and 's' and so on.
2) you can use syntax like:
Sentence[i], sentence[len(sentence)-i] = sentence[len(sentence)-i], Sentence[i]
This should definitely be enough to get you going.
You can do:
new_sentence = list()
sentence = list(raw_input(" "))
while sentence:
new_sentence.append(sentence.pop(-1))
else:
sentence = ''.join(new_sentence)
I have been learning Python (as my first language) from "How to Think Like a Computer Scientist: Learning with Python". This open book teaches mostly through examples and I prefer to read the goal and build the program on my own, rather than actually reading the program code provided in the book.
However, I am struggling with creating a function which will search for a specific character in a given string and return how many times that character was counted.
The code I wrote is:
def find(s, x): #find s in x
s = raw_input("Enter what you wish to find: ")
x = raw_input("Where to search? ")
count = 0
for l in x: #loop through every letter in x
if l == s:
count += 1
else:
print count
However, when I run this code, I get the error "name 's' is not defined".
The code in the book has a slightly different goal: it searches for a specific character in a string, but instead of counting how many times the character was found, it returns the position of the character in the string.
def find(strng, ch, start=0, step=1):
index = start
while 0 <= index < len(strng):
if strng[index] == ch:
return index
index += step
return -1
I don't really understand this code, actually.
However, even when I run the code, for example, to search for 'a' in 'banana', I get the error name 'banana' is not defined.
What is wrong with my code? Could please someone explain me how the code provided in the book works?
1: There are a couple things wrong with this code. The function takes in two parameters, s and x, then immediately throws them away by overwriting those variables with user input. In your for loop, every time you encounter a character that isn't s you print the count. You should try to separate different ideas in your code into different methods so that you can reuse code more easily.
Break down your code into small, simple ideas. If the purpose of find is to count the instances of a character in a string, it shouldn't also be handling user interaction. If you take out the raw_input and printing, you can simplify this function to:
def find(s, x): #find s in x
count = 0
for l in x: #loop through every letter in x
if l == s:
count += 1
return count
Now all it does it take in a character and a string and return the number of times the character appears in the string.
Now you can do your user interaction outside of the function
char = raw_input("Enter what you wish to find: ")
string = raw_input("Where to search?: )
print char + " appears " + `find(char, string)` + " times in " + string
2: The goal of this function is to find the first place where ch is found when walking through the characters strng from a starting position with a specified step. It takes in ch, strng, a position to start searching, and a step size. If the start is 0 and the step is 1, it will check every character. If the start is 2 it will check all but the first 2 characters, if the step is 2 it will check every other character, etc. This works by starting looking at the start index (index = start), then looping while the index is at least 0 and less than the length of the string. Since python is 0-indexed, the last character in the string has an index of one less than the length of the string, so this just restricts you from trying to check invalid indices. For each iteration of the loop, the code checks if the character at the current index is ch, in which case it returns the index (this is the first time it found the character). Every time it doesn't find the character at the current index, it increments the index by the step and tries again until it goes past the last character. When this happens it exits the loop and returns -1, a sentinel value which indicates that we didn't find the character in the string.
def find(strng, ch, start=0, step=1):
index = start
while 0 <= index < len(strng):
if strng[index] == ch:
return index
index += step
return -1
3: I'm guessing you passed some invalid parameters. strng should be a string, ch should be a single character, and start and step should be integers.
Try this. I took the parameters out of your function, moved the print command out of the else block and out of the for loop, and then wrote the last line to call the function.
def find(): #find s in x
s = raw_input("Enter what you wish to find: ")
x = raw_input("Where to search? ")
count = 0
for l in x: #loop through every letter in x
if l == s:
count += 1
print count
find()
It seems like you're taking in inputs s and x twice - once through the function arguments and once through raw input. Modify the function to do either one (say only from raw input - see below). Also, you only need to print out the count once, so you can place the print statement in the outermost indent level in the function.
def find(): #find s in x
s = raw_input("Enter what you wish to find: ")
x = raw_input("Where to search? ")
count = 0
for l in x: #loop through every letter in x
if l == s:
count += 1
print count