I need to keep my data in a list of lists, but want to edit elements in the lists based on their overall position.
For instance:
mylist = [['h','e','l','l','o'], ['w','o','r','l','d']]
I want to change position 5 as if it was all one list resulting in:
[['h','e','l','l','o'], ['change','o','r','l','d']]
This is for very large lists and lots of mutations so speed is essential!
Here is the solution of your's question
# initializing list
input_list = [['h', 'e', 'l', 'l', 'o'], ['w', 'o', 'r', 'l', 'd']]
print("The initial list is : " + str(input_list))
length: int = 0
# define position(global index) where you want to update items in list
change_index = 5
## global starting index of sublist [0,5]
sub_list_start_index: list = list()
for sub_list in input_list:
sub_list_start_index.append(length)
length += len(sub_list)
# check if index we want to change is <= global list index and
if change_index <= length - 1 and change_index >= max(sub_list_start_index):
sub_list_index = int(change_index - max(sub_list_start_index))
input_list[input_list.index(sub_list)][sub_list_index] = 'change'
print("Updated list : " + str(input_list))
Output:
The initial list is : [['h', 'e', 'l', 'l', 'o'], ['w', 'o', 'r', 'l', 'd']]
Updated list : [['h', 'e', 'l', 'l', 'o'], ['change', 'o', 'r', 'l', 'd']]
If I have two lists of individual string characters:
['H', 'e', 'l', 'l', 'o']
['w', 'o', 'r', 'l', 'd']
How do I make the final outcome look like this below of one list and where all of the string characters are combined:
['Hello','world']
If I try something like this:
word1join = "".join(word1)
word2join = "".join(word2)
print(word1join,type(word1join))
print(word2join,type(word1join))
print(list(word1join + word2join))
I am recreating the original data structure again but incorrectly, any tips appreciated!
Hello <class 'str'>
world <class 'str'>
['H', 'e', 'l', 'l', 'o', 'w', 'o', 'r', 'l', 'd']
Just create a new list with them in:
word1 = ['H', 'e', 'l', 'l', 'o']
word2 = ['w', 'o', 'r', 'l', 'd']
word1join = "".join(word1)
word2join = "".join(word2)
print([word1join, word2join])
Output as requested
If your variable names have numbers in them, it's usually a good hint that you should better be manipulating one list, rather than several individual variables.
Instead of starting with two variables:
word1 = ['H', 'e', 'l', 'l', 'o']
word2 = ['w', 'o', 'r', 'l', 'd']
Start with one list:
words = [['H', 'e', 'l', 'l', 'o'], ['w', 'o', 'r', 'l', 'd']]
You can still access individual words using indexing:
print(words[1])
# ['H', 'e', 'l', 'l', 'o']
You can then apply operations to every element of that list using for-loops, or list comprehensions, or function map:
words_joined = [''.join(word) for word in words]
# ['Hello', 'world']
# OR ALTERNATIVELY
words_joined = list(map(''.join, words))
# ['Hello', 'world']
If you really need to split the list into two variables, you can still do it:
word1joined, word2joined = words_joined
print(word2joined)
# world
I want to iterate over a list of characters
temp = ['h', 'e', 'l', 'l', 'o', '#', 'w', 'o', 'r', 'l', 'd']
so that I can obtain two strings, "hello" and "world"
My current way to do this is:
#temp is the name of the list
#temp2 is the starting index of the first alphabetical character found
for j in range(len(temp)):
if temp[j].isalpha() and temp[j-1] != '#':
temp2 = j
while (temp[temp2].isalpha() and temp2 < len(temp)-1:
temp2 += 1
print(temp[j:temp2+1])
j = temp2
The issue is that this prints out
['h', 'e', 'l', 'l', 'o']
['e', 'l', 'l', 'o']
['l', 'l', 'o']
['l', 'o']
['o']
etc. How can I print out only the full valid string?
Edit: I should have been more specific about what constitutes a "valid" string. A string is valid as long as all characters within it are either alphabetical or numerical. I didn't include the "isnumerical()" method within my check conditions because it isn't particularly relevant to the question.
If you want only hello and world and your words are always # seperated, you can easily do it by using join and split
>>> temp = ['h', 'e', 'l', 'l', 'o', '#', 'w', 'o', 'r', 'l', 'd']
>>> "".join(temp).split('#')
['hello', 'world']
Further more if you need to print the full valid string you need to
>>> t = "".join(temp).split('#')
>>> print(' '.join(t))
hello world
You can do it like this:
''.join(temp).split('#')
List has the method index which returns position of an element. You can use slicing to join the characters.
In [10]: temp = ['h', 'e', 'l', 'l', 'o', '#', 'w', 'o', 'r', 'l', 'd']
In [11]: pos = temp.index('#')
In [14]: ''.join(temp[:pos])
Out[14]: 'hello'
In [17]: ''.join(temp[pos+1:])
Out[17]: 'world'
An alternate, itertools-based solution:
>>> temp = ['h', 'e', 'l', 'l', 'o', '#', 'w', 'o', 'r', 'l', 'd']
>>> import itertools
>>> ["".join(str)
for isstr, str in itertools.groupby(temp, lambda c: c != '#')
if isstr]
['hello', 'world']
itertools.groupby is used to ... well ... group consecutive items depending if they are of not equal to #. The comprehension list will discard the sub-lists containing only # and join the non-# sub-lists.
The only advantage is that way, you don't have to build the full-string just to split it afterward. Probably only relevant if the string in really long.
If you just want alphas just use isalpha() replacing the # and any other non letters with a space and then split of you want a list of words:
print("".join(x if x.isalpha() else " " for x in temp).split())
If you want both words in a single string replace the # with a space and join using the conditional expression :
print("".join(x if x.isalpha() else " " for x in temp))
hello world
To do it using a loop like you own code just iterate over items and add to the output string is isalpha else add a space to the output:
out = ""
for s in temp:
if s.isalpha():
out += s
else:
out += " "
Using a loop to get a list of words:
words = []
out = ""
for s in temp:
if s.isalpha():
out += s
else:
words.append(out)
out = ""
This question already has answers here:
Strange result when removing item from a list while iterating over it
(8 answers)
Closed 7 years ago.
in this code I am trying to create a function anti_vowel that will remove all vowels (aeiouAEIOU) from a string. I think it should work ok, but when I run it, the sample text "Hey look Words!" is returned as "Hy lk Words!". It "forgets" to remove the last 'o'. How can this be?
text = "Hey look Words!"
def anti_vowel(text):
textlist = list(text)
for char in textlist:
if char.lower() in 'aeiou':
textlist.remove(char)
return "".join(textlist)
print anti_vowel(text)
You're modifying the list you're iterating over, which is bound to result in some unintuitive behavior. Instead, make a copy of the list so you don't remove elements from what you're iterating through.
for char in textlist[:]: #shallow copy of the list
# etc
To clarify the behavior you're seeing, check this out. Put print char, textlist at the beginning of your (original) loop. You'd expect, perhaps, that this would print out your string vertically, alongside the list, but what you'll actually get is this:
H ['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
e ['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!'] # !
l ['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
o ['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
k ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!'] # Problem!!
['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
W ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
o ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
d ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
s ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
! ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
Hy lk Words!
So what's going on? The nice for x in y loop in Python is really just syntactic sugar: it still accesses list elements by index. So when you remove elements from the list while iterating over it, you start skipping values (as you can see above). As a result, you never see the second o in "look"; you skip over it because the index has advanced "past" it when you deleted the previous element. Then, when you get to the o in "Words", you go to remove the first occurrence of 'o', which is the one you skipped before.
As others have mentioned, list comprehensions are probably an even better (cleaner, clearer) way to do this. Make use of the fact that Python strings are iterable:
def remove_vowels(text): # function names should start with verbs! :)
return ''.join(ch for ch in text if ch.lower() not in 'aeiou')
Other answers tell you why for skips items as you alter the list. This answer tells you how you should remove characters in a string without an explicit loop, instead.
Use str.translate():
vowels = 'aeiou'
vowels += vowels.upper()
text.translate(None, vowels)
This deletes all characters listed in the second argument.
Demo:
>>> text = "Hey look Words!"
>>> vowels = 'aeiou'
>>> vowels += vowels.upper()
>>> text.translate(None, vowels)
'Hy lk Wrds!'
>>> text = 'The Quick Brown Fox Jumps Over The Lazy Fox'
>>> text.translate(None, vowels)
'Th Qck Brwn Fx Jmps vr Th Lzy Fx'
In Python 3, the str.translate() method (Python 2: unicode.translate()) differs in that it doesn't take a deletechars parameter; the first argument is a dictionary mapping Unicode ordinals (integer values) to new values instead. Use None for any character that needs to be deleted:
# Python 3 code
vowels = 'aeiou'
vowels += vowels.upper()
vowels_table = dict.fromkeys(map(ord, vowels))
text.translate(vowels_table)
You can also use the str.maketrans() static method to produce that mapping:
vowels = 'aeiou'
vowels += vowels.upper()
text.translate(text.maketrans('', '', vowels))
Quoting from the docs:
Note: There is a subtlety when the sequence is being modified by the
loop (this can only occur for mutable sequences, i.e. lists). An
internal counter is used to keep track of which item is used next, and
this is incremented on each iteration. When this counter has reached
the length of the sequence the loop terminates. This means that if the
suite deletes the current (or a previous) item from the sequence, the
next item will be skipped (since it gets the index of the current item
which has already been treated). Likewise, if the suite inserts an
item in the sequence before the current item, the current item will be
treated again the next time through the loop. This can lead to nasty
bugs that can be avoided by making a temporary copy using a slice of
the whole sequence, e.g.,
for x in a[:]:
if x < 0: a.remove(x)
Iterate over a shallow copy of the list using [:]. You're modifying a list while iterating over it, this will result in some letters being missed.
The for loop keeps track of index, so when you remove an item at index i, the next item at i+1th position shifts to the current index(i) and hence in the next iteration you'll actually pick the i+2th item.
Lets take an easy example:
>>> text = "whoops"
>>> textlist = list(text)
>>> textlist
['w', 'h', 'o', 'o', 'p', 's']
for char in textlist:
if char.lower() in 'aeiou':
textlist.remove(char)
Iteration 1 : Index = 0.
char = 'W' as it is at index 0. As it doesn't satisfies that condition you'll do noting.
Iteration 2 : Index = 1.
char = 'h' as it is at index 1. Nothing more to do here.
Iteration 3 : Index = 2.
char = 'o' as it is at index 2. As this item satisfies the condition so it'll be removed from the list and all the items to it's right will shift one place to the left to fill the gap.
now textlist becomes :
0 1 2 3 4
`['w', 'h', 'o', 'p', 's']`
As you can see the other 'o' moved to index 2, i.e the current index so it'll be skipped in the next iteration. So, this is the reason some items are bring skipped in your iteration. Whenever you remove an item the next item is skipped from the iteration.
Iteration 4 : Index = 3.
char = 'p' as it is at index 3.
....
Fix:
Iterate over a shallow copy of the list to fix this issue:
for char in textlist[:]: #note the [:]
if char.lower() in 'aeiou':
textlist.remove(char)
Other alternatives:
List comprehension:
A one-liner using str.join and a list comprehension:
vowels = 'aeiou'
text = "Hey look Words!"
return "".join([char for char in text if char.lower() not in vowels])
regex:
>>> import re
>>> text = "Hey look Words!"
>>> re.sub('[aeiou]', '', text, flags=re.I)
'Hy lk Wrds!'
You're modifying the data you're iterating over. Don't do that.
''.join(x for x in textlist in x not in VOWELS)
text = "Hey look Words!"
print filter(lambda x: x not in "AaEeIiOoUu", text)
Output
Hy lk Wrds!
You're iterating over a list and deleting elements from it at the same time.
First, I need to make sure you clearly understand the role of char in for char in textlist: .... Take the situation where we have reached the letter 'l'. The situation is not like this:
['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
char
There is no link between char and the position of the letter 'l' in the list. If you modify char, the list will not be modified. The situation is more like this:
['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
char = 'l'
Notice that I've kept the ^ symbol. This is the hidden pointer that the code managing the for char in textlist: ... loop uses to keep track of its position in the loop. Every time you enter the body of the loop, the pointer is advanced, and the letter referenced by the pointer is copied into char.
Your problem occurs when you have two vowels in succession. I'll show you what happens from the point where you reach 'l'. Notice that I've also changed the word "look" to "leap", to make it clearer what's going on:
advance pointer to next character ('l') and copy to char
['H', 'e', 'y', ' ', 'l', 'e', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
-> ^
char = 'l'
char ('l') is not a vowel, so do nothing
advance pointer to next character ('e') and copy to char
['H', 'e', 'y', ' ', 'l', 'e', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
-> ^
char = 'e'
char ('e') is a vowel, so delete the first occurrence of char ('e')
['H', 'e', 'y', ' ', 'l', 'e', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
['H', 'e', 'y', ' ', 'l', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
['H', 'e', 'y', ' ', 'l', <- 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
['H', 'e', 'y', ' ', 'l', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
advance pointer to next character ('p') and copy to char
['H', 'e', 'y', ' ', 'l', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
-> ^
char = 'p'
When you removed the 'e' all the characters after the 'e' moved one place to the left, so it was as if remove had advanced the pointer. The result is that you skipped past the 'a'.
In general, you should avoid modifying lists while iterating over them. It's better to construct a new list from scratch, and Python's list comprehensions are the perfect tool for doing this. E.g.
print ''.join([char for char in "Hey look Words" if char.lower() not in "aeiou"])
But if you haven't learnt about comprehensions yet, the best way is probably:
text = "Hey look Words!"
def anti_vowel(text):
textlist = list(text)
new_textlist = []
for char in textlist:
if char.lower() not in 'aeiou':
new_textlist.append(char)
return "".join(new_textlist)
print anti_vowel(text)
List Comprehensions:
vowels = 'aeiou'
text = 'Hey look Words!'
result = [char for char in text if char not in vowels]
print ''.join(result)
Others have already explained the issue with your code. For your task, a generator expression is easier and less error prone.
>>> text = "Hey look Words!"
>>> ''.join(c for c in text if c.lower() not in 'aeiou')
'Hy lk Wrds!'
or
>>> ''.join(c for c in text if c not in 'AaEeIiOoUu')
'Hy lk Wrds!'
however, str.translate is the best way to go.
You shouldn't delete items from list you iterating through:
But you can make new list from the old one with list comprehension syntax. List comprehension is very useful in this situation. You can read about list comprehension here
So you solution will look like this:
text = "Hey look Words!"
def anti_vowel(text):
return "".join([char for char in list(text) if char.lower() not in 'aeiou'])
print anti_vowel(text)
It's pretty, isn't it :P
Try to not use the list() function on a string. It will make things a lot more complicated.
Unlike Java, in Python, strings are considered as arrays. Then, try to use an index for loop and del keyword.
for x in range(len(string)):
if string[x].lower() in "aeiou":
del string[x]
def make_str_from_row(board, row_index):
''' (list of list of str, int) -> str
Return the characters from the row of the board with index row_index
as a single string.
>>> make_str_from_row([['A', 'N', 'T', 'T'], ['X', 'S', 'O', 'B']], 0)
'ANTT'
'''
for i in range(len(board)):
i = row_index
print(board[i])
This prints ['A', 'N', 'T', 'T']
How do I print it like this 'ANTT' instead?
Well, you got what you told to print!
board is a list of list of strs, so board[i] must be a list of strs, and when you write print(board[i]), you get a list!
You may need to write this:
print(''.join(board[i]))
You could simplify that a whole lot by using
>>> def make_str_from_row(board, row_index):
... print repr(''.join(board[row_index]))
...
>>> make_str_from_row([['A', 'N', 'T', 'T'], ['X', 'S', 'O', 'B']], 0)
'ANTT'
The reason you get that output is because you print a list since the elements of board are lists. By using join, you get a string.
Also, I don't understand why you use a loop if are going to change the index you loop over.
I think this was what you were trying to do:
def make_str_from_row(board, row_index):
''' (list of list of str, int) -> str
Return the characters from the row of the board with index row_index
as a single string.
>>> make_str_from_row([['A', 'N', 'T', 'T'], ['X', 'S', 'O', 'B']], 0)
'ANTT'
'''
for cell in board[row_index]:
print cell,