I thought that this code below could remove the trailing character of every element in a. I can't understand the output
a = ['wonderland#', 'alice,', 'in!$', 'book:']
for index, word in enumerate(a):
for ch in '#!$,:':
a[index] = word.strip(ch)
print(a)
>>>['wonderland#', 'alice,', 'in!$', 'book']
Recommended:
Try rstrip this way with a list comprehension:
a = ['wonderland#', 'alice,', 'in!$', 'book:']
print([i.rstrip('#!$,:') for i in a])
Or try regex:
import re
print([re.sub('[#!$,:]+$', '', i) for i in a])
Both codes output:
['wonderland', 'alice', 'in', 'book']
Not recommended:
The reason your code doesn't work is because you may well matched the character(s) at the end but the next iteration you would be still trying to strip from the original string, instead of the string that you just stripped in the previous iterations. Also you use strip so it might be that only one character would be striped, so you also need to use sorted, so to fix your code:
a = ['wonderland#', 'alice,', 'in!$', 'book:']
for index, word in enumerate(a):
for ch in sorted('#!$,:', key=word.find)[::-1]:
a[index] = a[index].strip(ch)
print(a)
print(a)
Output:
['wonderland', 'alice', 'in', 'book']
The problem is that you're iterating over the list and assigning it to word, then every call to strip starts over with the unmodified word. Try this instead:
for index in range(len(a)):
for ch in '#!$,:':
a[index] = a[index].strip(ch)
And it turns out that doesn't work either, because it's dependent on the order of the characters that you're removing. Take advantage of strips ability to remove multiple characters at once.
for index in range(len(a)):
a[index] = a[index].strip('#!$,:')
Related
I have a list like this:
my_list=["'-\\n'",
"'81\\n'",
"'-\\n'",
"'0913\\n'",
"'Assistant nursing\\n'",
"'0533\\n'",
"'0895 Astronomy\\n'",
"'0533\\n'",
"'Astrophysics\\n'",
"'0532\\n'"]
Is there any way to delete every thing from this list except words?
out put:
my_list=['Assistant nursing',
'Astronomy',
'Astrophysics',]
I know for example if i wanna remove integers in string form i can do this:
no_integers = [x for x in my_list if not (x.isdigit()
or x[0] == '-' and x[1:].isdigit())]
but it dosn't work well enough
The non-regex solution:
You can start by striping off the characters '-\\n, then take only the characters that are alphabets using str.isalpha or a white space, then filter out the sub-strings that are empty ''. You may need to strip off the white space characters in the end, whic
>>> list(filter(lambda x: x!='', (''.join(j for j in i.strip('\'-\\\\n') if j.isalpha() or j==' ').strip() for i in my_list)))
['Assistant nursing', 'Astronomy', 'Astrophysics']
If you want to use regex, you can use the pattern: '([A-Za-z].*?)\\\\n' with re.findall, then filter out the elements that are empty list, finally you can flatten the list
>>> import re
>>> list(filter(lambda x: x, [re.findall('([A-Za-z].*?)\\\\n', i) for i in my_list]))
[['Assistant nursing'], ['Astronomy'], ['Astrophysics']]
with regular expresssions
import re
my_list = # above
# remove \n, -, digits, ' symbols
my_new_list = [re.sub(r"[\d\\n\-']", '', s) for s in my_list]
# remove empty strings
my_new_list = [s for s in my_new_list if s != '']
print(my_new_list)
Output
['Assistat ursig', ' Astroomy', 'Astrophysics']
Input a given string and check if any word in that string matches with its reverse in the same string then print that word else print $
I split the string and put the words in a list and then I reversed the words in that list. After that, I couldn't able to compare both the lists.
str = input()
x = str.split()
for i in x: # printing i shows the words in the list
str1 = i[::-1] # printing str1 shows the reverse of words in a new list
# now how to check if any word of the new list matches to any word of the old list
if(i==str):
print(i)
break
else:
print('$)
Input: suman is a si boy.
Output: is ( since reverse of 'is' is present in the same string)
You almost have it, just need to add another loop to compare each word against each inverted word. Try using the following
str = input()
x = str.split()
for i in x:
str1 = i[::-1]
for j in x: # <-- this is the new nested loop you are missing
if j == str1: # compare each inverted word against each regular word
if len(str1) > 1: # Potential condition if you would like to not include single letter words
print(i)
Update
To only print the first occurrence of a match, you could, in the second loop, only check the elements that come after. We can do this by keeping track of the index:
str = input()
x = str.split()
for index, i in enumerate(x):
str1 = i[::-1]
for j in x[index+1:]: # <-- only consider words that are ahead
if j == str1:
if len(str1) > 1:
print(i)
Note that I used index+1 in order to not consider single word palindromes a match.
a = 'suman is a si boy'
# Construct the list of words
words = a.split(' ')
# Construct the list of reversed words
reversed_words = [word[::-1] for word in words]
# Get an intersection of these lists converted to sets
print(set(words) & set(reversed_words))
will print:
{'si', 'is', 'a'}
Another way to do this is just in a list comprehension:
string = 'suman is a si boy'
output = [x for x in string.split() if x[::-1] in string.split()]
print(output)
The split on string creates a list split on spaces. Then the word is included only if the reverse is in the string.
Output is:
['is', 'a', 'si']
One note, you have a variable name str. Best not to do that as str is a Python thing and could cause other issues in your code later on.
If you want word more than one letter long then you can do:
string = 'suman is a si boy'
output = [x for x in string.split() if x[::-1] in string.split() and len(x) > 1]
print(output)
this gives:
['is', 'si']
Final Answer...
And for the final thought, in order to get just the 'is':
string = 'suman is a si boy'
seen = []
output = [x for x in string.split() if x[::-1] not in seen and not seen.append(x) and x[::-1] in string.split() and len(x) > 1]
print(output)
output is:
['is']
BUT, this is not necessarily a good way to do it, I don't believe. Basically you are storing information in seen during the list comprehension AND referencing that same list. :)
This answer wouldn't show you 'a' and won't output 'is' with 'si'.
str = input() #get input string
x = str.split() #returns list of words
y = [] #list of words
while len(x) > 0 :
a = x.pop(0) #removes first item from list and returns it, then assigns it to a
if a[::-1] in x: #checks if the reversed word is in the list of words
#the list doesn't contain that word anymore so 'a' that doesn't show twice wouldn't be returned
#and 'is' that is present with 'si' will be evaluated once
y.append(a)
print(y) # ['is']
I have a very messy data, I am trying to remove elements that contains alphabets or words. I am trying to capture the elements that have alphanumerical and numerical values. I tried .isalpha() but it not working. How do I remove this?
lista = ['A8817-2938-228','12421','12323-12928-A','12323-12928',
'-','A','YDDEWE','hello','world','testing_purpose','testing purpose',
'A8232-2938-228','N7261-8271']
lista
Tried:
[i.isalnum() for i in lista] # gives boolean, but opposite of what I need.
Output:
['A8817-2938-228','12421','12323-12928-A','12323-12928','-','A8232-2938-228','N7261-8271']
Thanks!
You can add conditional checks in list comprehensions, so this is what you want:
new_list = [i for i in lista if not i.isalnum()]
print(new_list)
Output:
['A8817-2938-228', '12323-12928-A', '12323-12928', '-', 'testing_purpose', 'testing purpose', 'A8232-2938-228', 'N7261-8271']
Note that isalnum won't say True if the string contains spaces or underscores. One option is to remove them before checking: (You also need to use isalpha instead of isalnum)
new_list_2 = [i for i in lista if not i.replace(" ", "").replace("_", "").isalpha()]
print(new_list_2)
Output:
['A8817-2938-228', '12421', '12323-12928-A', '12323-12928', '-', 'A8232-2938-228', 'N7261-8271']
It seems you can just test at least one character is a digit or equality with '-':
res = [i for i in lista if any(ch.isdigit() for ch in i) or i == '-']
print(res)
['A8817-2938-228', '12421', '12323-12928-A', '12323-12928',
'-', 'A8232-2938-228', 'N7261-8271']
What type your data in the list?
You can try to do this:
[str(i).isalnum() for i in lista]
I have following data in a list and it is a hex number,
['aaaaa955554e']
I would like to split this into ['aaaaa9,55554e'] with a comma.
I know how to split this when there are some delimiters between but how should i do for this case?
Thanks
This will do what I think you are looking for:
yourlist = ['aaaaa955554e']
new_list = [','.join([x[i:i+6] for i in range(0, len(x), 6)]) for x in yourlist]
It will put a comma at every sixth character in each item in your list. (I am assuming you will have more than just one item in the list, and that the items are of unknown length. Not that it matters.)
i assume you wanna split into every 6th character
using regex
import re
lst = ['aaaaa955554e']
newlst = re.findall('\w{6}', lst[0])
# ['aaaaa9', '55554e']
Using list comprehension, this works for multiple items in lst
lst = ['aaaaa955554e']
newlst = [item[i:i+6] for i in range(0,len(a[0]),6) for item in lst]
# ['aaaaa9', '55554e']
This could be done using a regular expression substitution as follows:
import re
print re.sub(r'([a-zA-Z]+\d)(.*?)', r'\1,\2', 'aaaaa955554e', count=1)
Giving you:
aaaaa9,55554e
This splits after seeing the first digit.
s = ['my', 'name']
I want to change the 1st letter of each element in to Upper Case.
s = ['My', 'Name']
Both .capitalize() and .title(), changes the other letters in the string to lower case.
Here is a simple function that only changes the first letter to upper case, and leaves the rest unchanged.
def upcase_first_letter(s):
return s[0].upper() + s[1:]
You can use the capitalize() method:
s = ['my', 'name']
s = [item.capitalize() for item in s]
print s # print(s) in Python 3
This will print:
['My', 'Name']
You can use 'my'.title() which will return 'My'.
To get over the complete list, simply map over it like this:
>>> map(lambda x: x.title(), s)
['My', 'Name']
Actually, .title() makes all words start with uppercase. If you want to strictly limit it the first letter, use capitalize() instead. (This makes a difference for example in 'this word' being changed to either This Word or This word)
It probably doesn't matter, but you might want to use this instead of the capitalize() or title() string methods because, in addition to uppercasing the first letter, they also lowercase the rest of the string (and this doesn't):
s = map(lambda e: e[:1].upper() + e[1:] if e else '', s)
Note: In Python 3, you'd need to use:
s = list(map(lambda e: e[:1].upper() + e[1:] if e else '', s))
because map() returns an iterator that applies function to every item of iterable instead of a list as it did in Python 2 (so you have to turn it into one yourself).
You can use
for i in range(len(s)):
s[i]=s[i].capitalize()
print s