i've searched the forum and found similar questions, but no luck in solving my problem.
My code is designed to swap every two letters of each word using recursion and print the result. For words with an even amount of letters, the word "None" is included in the output and i don't know how to fix...
here's the code:
def encryptLine(line, count):
headline = line[count:]
if length(headline) > 0:
if count == length(line) - 1:
new = headline
return new
elif count <= length(line):
new = head(tail(headline)) + head(headline)
new = new + str(encryptLine(line, count+2))
return new
print(encryptLine('abcd', 0))
the output for 'abcd' is badcNone, which is correct except for the word None. the output for 'abcde' is 'badce', which is correct...
thanks in advance for your help!
Add return "" to the function definition, that is
def encryptLine(line, count):
headline = line[count:]
if length(headline) > 0:
if count == length(line) - 1:
new = headline
return new
elif count <= length(line):
new = head(tail(headline)) + head(headline)
new = new + str(encryptLine(line, count+2))
return new
return ""
Otherwise, the function will return None if length(headline) > 0 does not hold.
None is here because your function return nothing.
There is 1 case where you return nothing it is
if length(headline) <= 0:
In Python, if there is no return to a function and you try to access to a return value, the value will be None.
Related
Here in this code else block is not printing the value Treasure locked
def counted(value):
if(value == 5):
return 1
else:
return 0
def numb(value1):
sam = 95
value = 0
stp = 97
h = {}
for i in range(0,26):
h.update({chr(stp) : (ord(chr(stp))-sam)})
sam = sam-1
stp = stp+1
for j in range(0,5):
value = h[value1[j]]+value
if(value > 80):
print('First lock-unlocked')
else:
print('Treasure locked')
string = input()
firstcheck = counted(len(string))
if(firstcheck == 1):
numb(string)
a good idea is to check what the condition is before entering the if statements, possibly check what value is printing before the if statement. the logic in def numb() has very little do with what's in def counted(). as long as one is 1 or 0 is being passed to numb() we know that function will run and seems like it.
else block is working properly. if you want to print Treasure Locked. you have to pass lower character string like 'aaaaa'. if value is > 80. then it always print First lock-unlocked.
I created a HTML text cleaner, which deletes data between tags.
It's working fine on one iteration, but not in a loop.
The problem is, I cannot save newhtml as a variable due to Python's string immutability.
So, my loop is only working for the last iteration of the function return.
What would be the best practice in such a situation?
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += len(sub) # use start += 1 to find overlapping matches
def replace_string(index1, index2, mainstring):
replacementstring = ''
return mainstring.replace(mainstring[index1:index2], replacementstring)
def strip_images(html):
begin_indexes = list(find_all(html, '<DESCRIPTION>GRAPHIC'))
end_indexes = list(find_all(html, '</TEXT>'))
for i in range(len(begin_indexes)):
if begin_indexes[i] > end_indexes[i]:
end_indexes.pop(0)
else:
if len(begin_indexes) == len(end_indexes):
break
for i in range(len(begin_indexes)):
#code problem is here--
newhtml = replace_string(begin_indexes[i],end_indexes[i], html)
if i == len(begin_indexes) - 1:
return newhtml
#code only returns one iteration
var = strip_images(html)
print var
Your current issue is that html never changes within the loop. So, your input will always be for the first iteration, regardless of the length of the lists.
The solution here follows these steps
Assign the string to the original value before the loop
Edit within the loop, passing in the current content, returning a replaced string
Return from the function after the loop
newhtml = html
for begin, end in zip(begin_indexes, end_indexes):
newhtml = replace_string(begin, end, newhtml)
return newhtml
Got it working, here's the code snippet. It's not pretty but it's doing the job of removing text between those two tags:
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += len(sub) # use start += 1 to find overlapping matches
def strip_images(html):
begin_indexes = list(find_all(html, '<DESCRIPTION>GRAPHIC'))
end_indexes = list(find_all(html, '</TEXT>'))
for i in range(len(begin_indexes)):
if begin_indexes[i] > end_indexes[i]:
end_indexes.pop(0)
else:
if len(begin_indexes) == len(end_indexes):
break
newhtml = html
begin_indexes2 = begin_indexes[::-1]
end_indexes2 = end_indexes[::-1]
for i in range(len(begin_indexes2)):
#for i, value in enumerate(begin_indexes,0):
#end_indexes.reset_index(drop=True)
newhtml = list(newhtml)
del newhtml[begin_indexes2[i]:end_indexes2[i]]
if i == len(begin_indexes2) - 1:
str1 = ''.join(newhtml)
return str1
I am new to this, and I am looking for help. I currently am stuck in a program I'm trying to complete. Here it is:
def searchStock(stockList, stockPrice, s):
for i in range(len(stockList)):
if s == stockList[i]:
s = stockPrice[i]
elif s != stockList[i]:
s = -1
return s
def mainFun():
stockList= []
stockPrice = []
l = 1
while l > 0:
stocks = str(input("Enter the name of the stock:"))
stockList += [stocks]
if stocks == "done"or stocks == 'done':
l = l * -1
stockList.remove("done")
else:
price = int(input("Enter the price of the stock:"))
stockPrice += [price]
l = l + 1
print(stockList)
print(stockPrice)
s = input("Enter the name of the stock you're looking for:")
s = searchStock(stockList, stockPrice, s)
Every time I run the program to the end, it never returns the variable s for some reason. If i replace return with print, it always prints -1 instead of the stockPrice if its on the list. I cant seem to get it to work. Can someone please help me?
Try adding this print to help you debug:
def searchStock(stockList, stockPrice, s):
output = -1
for i in range(len(stockList)):
if s == stockList[i]:
output = stockPrice[i]
print i, output, stockList[i], stockPrice[i]
elif s != stockList[i]:
output = -1
return output
Also I changed one of your variables, it seems better than modifying your input value and then returning it.
I want to do something like this:
>>> mystring = "foo"
>>> print(mid(mystring))
Help!
slices to the rescue :)
def left(s, amount):
return s[:amount]
def right(s, amount):
return s[-amount:]
def mid(s, offset, amount):
return s[offset:offset+amount]
If I remember my QBasic, right, left and mid do something like this:
>>> s = '123456789'
>>> s[-2:]
'89'
>>> s[:2]
'12'
>>> s[4:6]
'56'
http://www.angelfire.com/scifi/nightcode/prglang/qbasic/function/strings/left_right.html
Thanks Andy W
I found that the mid() did not quite work as I expected and I modified as follows:
def mid(s, offset, amount):
return s[offset-1:offset+amount-1]
I performed the following test:
print('[1]23', mid('123', 1, 1))
print('1[2]3', mid('123', 2, 1))
print('12[3]', mid('123', 3, 1))
print('[12]3', mid('123', 1, 2))
print('1[23]', mid('123', 2, 2))
Which resulted in:
[1]23 1
1[2]3 2
12[3] 3
[12]3 12
1[23] 23
Which was what I was expecting. The original mid() code produces this:
[1]23 2
1[2]3 3
12[3]
[12]3 23
1[23] 3
But the left() and right() functions work fine. Thank you.
You can use this method also it will act like that
thadari=[1,2,3,4,5,6]
#Front Two(Left)
print(thadari[:2])
[1,2]
#Last Two(Right)# edited
print(thadari[-2:])
[5,6]
#mid
mid = len(thadari) //2
lefthalf = thadari[:mid]
[1,2,3]
righthalf = thadari[mid:]
[4,5,6]
Hope it will help
This is Andy's solution. I just addressed User2357112's concern and gave it meaningful variable names. I'm a Python rookie and preferred these functions.
def left(aString, howMany):
if howMany <1:
return ''
else:
return aString[:howMany]
def right(aString, howMany):
if howMany <1:
return ''
else:
return aString[-howMany:]
def mid(aString, startChar, howMany):
if howMany < 1:
return ''
else:
return aString[startChar:startChar+howMany]
These work great for reading left / right "n" characters from a string, but, at least with BBC BASIC, the LEFT$() and RIGHT$() functions allowed you to change the left / right "n" characters too...
E.g.:
10 a$="00000"
20 RIGHT$(a$,3)="ABC"
30 PRINT a$
Would produce:
00ABC
Edit : Since writing this, I've come up with my own solution...
def left(s, amount = 1, substring = ""):
if (substring == ""):
return s[:amount]
else:
if (len(substring) > amount):
substring = substring[:amount]
return substring + s[:-amount]
def right(s, amount = 1, substring = ""):
if (substring == ""):
return s[-amount:]
else:
if (len(substring) > amount):
substring = substring[:amount]
return s[:-amount] + substring
To return n characters you'd call
substring = left(string,<n>)
Where defaults to 1 if not supplied. The same is true for right()
To change the left or right n characters of a string you'd call
newstring = left(string,<n>,substring)
This would take the first n characters of substring and overwrite the first n characters of string, returning the result in newstring. The same works for right().
There are built-in functions in Python for "right" and "left", if you are looking for a boolean result.
str = "this_is_a_test"
left = str.startswith("this")
print(left)
> True
right = str.endswith("test")
print(right)
> True
Based on the comments above, it would seem that the Right() function could be refactored to handle errors better. This seems to work:
def right(s, amount):
if s == None:
return None
elif amount == None:
return None # Or throw a missing argument error
s = str(s)
if amount > len(s):
return s
elif amount == 0:
return ""
else:
return s[-amount:]
print(right("egg",2))
print(right(None, 2))
print(right("egg", None))
print(right("egg", 5))
print("a" + right("egg", 0) + "b")
I have an assignment said that to create a findString function that accept 2 string which are 'target' and 'query', and that returns
a
list
of
all
indices
in
target
where
query
appears.
If
target
does
not
contain
query,
return
an
empty
list.
For example:
findString(‘attaggtttattgg’,’gg’)
return:
[4,12]
I dont know how to start off with this function writing at all. Please help me everyone. Thank you so much!!!
since an answer has already been given:
def find_matches(strng, substrng):
substrg_len = len(substr)
return [i for i in range(len(strg) + 1 - substrg_len)
if strg[i:i+substrg_len] == substrg]
def find_string(search, needle):
start = -1
results = []
while start + 1< len(search):
start = search.find(needle, start +1)
if start == -1:
break
results.append(start )
return results
Here are a couple of hints to get you started.
target.find(query) will return the index of query in target. If query is not found, it will return -1.
A string can be sliced. target[pos:] will give you a substring of target starting from pos.
This may require some error handling:
def find_allPatterns(strVal, strSub):
listPos = []
strTemp = strVal
while True:
try:
posInStr = strTemp.index(strSub)
except ValueError:
posInStr = None
if posInStr:
listPos.append(posInStr)
subpos = posInStr + len(strSub)
strTemp = strTemp[subpos:]
else:
break
return listPos
print find_allPatterns('attaggtttattgg', 'gg')
Output:
[4, 6]