Re: Weird behavior in search in a list
"Su Y" <[EMAIL PROTECTED]> wrote: > I want find the first number in extend[] which is larger than num, so > I wrote: > def find(num): > count=0 > for elem in extend: > if elem count+=1 > return count > > I found that if extend[] is monotonous, like [1.1, 2.3, 3.2, 4.5, > 5.6], > it works fine: find(4) returns 3, extend[3] is 4.5. > But, if extend[] is not monotonous, like [1.1, 2.3, 3.2, 4.5, 5.6, > 4.6, 3.4, 2.1, 0.3], > find(4) returns 6, extend[6] is 3.4! > > what's going on here? I really can't understand Hint: extend[0] is1.1 Hint: extend[7] is 2.1 Hint: 2.1 is less than 4 You have to stop counting and come out of the loop when you find the first one - what your function is doing is counting the elements less than num, not finding the first one that is. hth - Hendrik -- http://mail.python.org/mailman/listinfo/python-list
Re: Weird behavior in search in a list
On 3月29日, 下午8时22分, Michael Bentley <[EMAIL PROTECTED]> wrote: > On Mar 29, 2007, at 6:51 AM, Su Y wrote: > > > > > > > I want find the first number in extend[] which is larger than num, so > > I wrote: > > def find(num): > > count=0 > > for elem in extend: > > if elem > count+=1 > > return count > > > I found that if extend[] is monotonous, like [1.1, 2.3, 3.2, 4.5, > > 5.6], > > it works fine: find(4) returns 3, extend[3] is 4.5. > > But, if extend[] is not monotonous, like [1.1, 2.3, 3.2, 4.5, 5.6, > > 4.6, 3.4, 2.1, 0.3], > > find(4) returns 6, extend[6] is 3.4! > > > what's going on here? I really can't understand > > find() loops through the list, and every time it finds a value less > than num it increments count. So in your second example the values > 1.1, 2.3, 3.2, 3.4, 2.1, and 0.3 are all less than 4, which means > count will be 6. > > As you learned, a sorted list behaves as you expect. So one approach > is to sort your list first. But another perhaps better approach is > to do something like: > > def find(num): > # check to make sure there *is* a value greater than num > if max(extend) > num: > # then return the smallest value that is greater than num > return min([x for x in extend if x > num]) > else: > return None > > Hope this helps, > Michael oh yes, it's the "break" I forgot... Thank you, it helps me a lot! -- http://mail.python.org/mailman/listinfo/python-list
Re: Weird behavior in search in a list
On 3/29/07, Michael Bentley <[EMAIL PROTECTED]> wrote: > > On Mar 29, 2007, at 6:51 AM, Su Y wrote: > > > > I want find the first number in extend[] which is larger than num, so > def find(num): > # check to make sure there *is* a value greater than num > if max(extend) > num: > # then return the smallest value that is greater than num > return min([x for x in extend if x > num]) > else: > return None I think OP wants the first value in the list greater than 'num' not the smallest greater value in the list. cheers, -- Amit Khemka -- onyomo.com Home Page: www.cse.iitd.ernet.in/~csd00377 Endless the world's turn, endless the sun's Spinning, Endless the quest; I turn again, back to my own beginning, And here, find rest. -- http://mail.python.org/mailman/listinfo/python-list
Re: Weird behavior in search in a list
On Mar 29, 2007, at 6:51 AM, Su Y wrote: > > I want find the first number in extend[] which is larger than num, so > I wrote: > def find(num): > count=0 > for elem in extend: > if elem count+=1 > return count > > I found that if extend[] is monotonous, like [1.1, 2.3, 3.2, 4.5, > 5.6], > it works fine: find(4) returns 3, extend[3] is 4.5. > But, if extend[] is not monotonous, like [1.1, 2.3, 3.2, 4.5, 5.6, > 4.6, 3.4, 2.1, 0.3], > find(4) returns 6, extend[6] is 3.4! > > what's going on here? I really can't understand find() loops through the list, and every time it finds a value less than num it increments count. So in your second example the values 1.1, 2.3, 3.2, 3.4, 2.1, and 0.3 are all less than 4, which means count will be 6. As you learned, a sorted list behaves as you expect. So one approach is to sort your list first. But another perhaps better approach is to do something like: def find(num): # check to make sure there *is* a value greater than num if max(extend) > num: # then return the smallest value that is greater than num return min([x for x in extend if x > num]) else: return None Hope this helps, Michael -- http://mail.python.org/mailman/listinfo/python-list
Re: Weird behavior in search in a list
"Su Y" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] On 3ÔÂ29ÈÕ, ÏÂÎç7ʱ51·Ö, "Su Y" <[EMAIL PROTECTED]> wrote: > hi all, > I can't understand how this code work, its behavior is really weird > for me... > > I want find the first number in extend[] which is larger than num, soI > wrote: > > def find(num): > count=0 > for elem in extend: > if elem count+=1 else: break > return count > you need to break out of the loop when you first encounter num>elem. The reason it works in your sorted list scenario is because elem will be > num, always, after some point. It won't be in your unsorted list. I've added the else: break in your code above j -- http://mail.python.org/mailman/listinfo/python-list
Re: Weird behavior in search in a list
> On 29 Mar 2007 04:51:00 -0700, Su Y <[EMAIL PROTECTED]> wrote: > > I want find the first number in extend[] which is larger than num, so > On 3/29/07, Amit Khemka <[EMAIL PROTECTED]> wrote: > Btw a concise way could be: > def getfirstbigger(num): > for i,x in enumerate(extend): > if x>num:return i > return len(extend) I forgot to add that you can as well 'return' the the item (along with the index), but in case all the items in the list are smaller than the argument, what return value do you require? -- Amit Khemka -- onyomo.com Home Page: www.cse.iitd.ernet.in/~csd00377 Endless the world's turn, endless the sun's Spinning, Endless the quest; I turn again, back to my own beginning, And here, find rest. -- http://mail.python.org/mailman/listinfo/python-list
Re: Weird behavior in search in a list
On 29 Mar 2007 04:51:00 -0700, Su Y <[EMAIL PROTECTED]> wrote: > hi all, > I can't understand how this code work, its behavior is really weird > for me... > > I want find the first number in extend[] which is larger than num, so > I wrote: > def find(num): > count=0 > for elem in extend: > if elem count+=1 > return count > > I found that if extend[] is monotonous, like [1.1, 2.3, 3.2, 4.5, > 5.6], > it works fine: find(4) returns 3, extend[3] is 4.5. > But, if extend[] is not monotonous, like [1.1, 2.3, 3.2, 4.5, 5.6, > 4.6, 3.4, 2.1, 0.3], > find(4) returns 6, extend[6] is 3.4! > > what's going on here? I really can't understand Actually your function "find" returns the number of elements in list extend, which are smaller than the argument. Perhaps you wanted to 'break' when once the condition is met. def find(num): count=0 for elem in extend: if elem>num: break else: count+=1 return count Btw a concise way could be: def getfirstbigger(num): for i,x in enumerate(extend): if x>num:return i return len(extend) HTH, -- Amit Khemka -- onyomo.com Home Page: www.cse.iitd.ernet.in/~csd00377 Endless the world's turn, endless the sun's Spinning, Endless the quest; I turn again, back to my own beginning, And here, find rest. -- http://mail.python.org/mailman/listinfo/python-list
Re: Weird behavior in search in a list
On 3月29日, 下午7时51分, "Su Y" <[EMAIL PROTECTED]> wrote: > hi all, > I can't understand how this code work, its behavior is really weird > for me... > > I want find the first number in extend[] which is larger than num, soI wrote: > > def find(num): > count=0 > for elem in extend: > if elem count+=1 > return count > > I found that if extend[] is monotonous, like [1.1, 2.3, 3.2, 4.5, > 5.6], > it works fine: find(4) returns 3, extend[3] is 4.5. > But, if extend[] is not monotonous, like [1.1, 2.3, 3.2, 4.5, 5.6, > 4.6, 3.4, 2.1, 0.3], > find(4) returns 6, extend[6] is 3.4! > > what's going on here? I really can't understand and I am using Python 2.5 on WinXP. -- http://mail.python.org/mailman/listinfo/python-list
Weird behavior in search in a list
hi all, I can't understand how this code work, its behavior is really weird for me... I want find the first number in extend[] which is larger than num, so I wrote: def find(num): count=0 for elem in extend: if elem-- http://mail.python.org/mailman/listinfo/python-list
