lines = []
while True:
line = raw_input()
if line:
lines.append(line)
else:
break
print lines
This would take line by line in a list. Output is:
In [27]: lines
Out[27]: ['x-xx', 'y->y', '-z->']
How do I access the next letter, currently being at a letter, in the following specified code:
count = 0 # to check how many '->' are there in each line
for sentence in lines:
for letter in sentence:
if letter == '-':
#check if the next character is '>' (How to code this line)
#and if so, increment count
else:
break
Is there a way out for this kind of for loop, where you don't index letter but iterate on letter itself directly?
Related
how to keep the space betwen the words?
in the code it deletes them and prints them in column.. so how to print them in row and with the space?
s ='[]'
f = open('q4.txt', "r")
for line in f:
for word in line:
b = word.strip()
c = list(b)
for j in b:
if ord(j) == 32:
print ord(33)
if ord(j) == 97:
print ord(123)
if ord(j) == 65:
print ord(91)
chr_nums = chr(ord(j) - 1)
print chr_nums
f.close()
Short answer: remove the word.strip() command - that's deleting the space. Then put a comma after the print operation to prevent a newline: print chr_nums,
There are several problems with your code aside from what you ask about here:
ord() takes a string (character) not an int, so ord(33) will fail.
for word in line: will be iterating over characters, not words, so word will be a single character and for j in b is unnecessary.
Take a look at the first for loop :
for line in f:
here the variable named 'line' is actually a line from the text file you are reading. So this 'line' variable is actually a string. Now take a look at the second for loop :
for word in line:
Here you are using a for loop on a string variable named as 'line' which we have got from the previous loop. So in the variable named 'word' you are not going to get a word, but single characters one by one. Let me demonstrate this using a simple example :
for word in "how are you?":
print(word)
The output of this code will be as follows :
h
o
w
a
r
e
y
o
u
?
You are getting individual characters from the line and so you don't need to use another for loop like you did 'for j in b:'. I hope this helped you.
I am writing a program that produce a frequency plot of the letters in a body of text. however, there is an error in my code that I can not spot it. any ideas?
def letter_count(word,freqs,pmarks):
for char in word:
freqs[char]+=1
def letter_freq(fname):
fhand = open(fname)
freqs = dict()
alpha = list(string.uppercase[:26])
for let in alpha: freqs[let] = freqs.get(let,0)
for line in fhand:
line = line.rstrip()
words = line.split()
pmarks = list(string.punctuation)
words = [word.upper() for word in words]
for word in words:
letter_count(word,freqs,pmarks)
fhand.close()
return freqs.values
You are calling
freqs[char]+=1
with char = '.' without having initialized a value freqs['.']=0
You should check before line 3, whether the key exists already, as you can do the +=1 operation only on existing keys of the dictionary.
So something like:
for char in word:
if freqs.has_key(char):
freqs[char]+=1
Python: how can I check if the key of an dictionary exists?
I'm making a hang man game. When I made the code with out a conditional and classes, it worked fine. Basically my issues with the code below are:
Only the letter "t" will match. I can't get any other letter to match.
If I enter "t" on the first try, then purposely get the next 4 letters wrong, it won't end until after 7 turns. Yet if I enter any other letter first, it will end after 4 wrong turns, like it should.
My questions....
How can I get it to match with the other letters that are in the self.word index?
Why is it not obeying the condition I set with the while loop in the main method if I enter "t" on my first try and get every other letter wrong thereafter?
class Hang():
def __init__(self):
self.turns = 0
self.word = ['t', 'h', 'i', 's']
self.empty = ["__", "__", "__", "__"]
self.wrong = []
def main(self):
while self.turns < 4:
for i in self.word:
choice = raw_input("Enter a letter a-z: ")
if choice == i:
index = self.word.index(i)
self.empty.pop(index)
self.empty.insert(index, i)
print self.empty
else:
print "Wrong"
self.wrong.append(choice)
print self.wrong
print self.empty
self.turns += 1
char1 = Hang()
char1.main()
In the game of hangman you can guess any character in the phrase in any order. But you're using a for loop to go through each character in order and it is only correct if the player correctly guesses the characters in order
Try this instead
while self.turns < 4:
choice = raw_input("Enter a letter a-z: ")
# optional, if user enters more than a single letter
if len(choice) > 1:
print "invalid choice"
continue # loop again from start
index = self.word.index(choice)
if index != -1:
# -1 indicates character in not int the string
# so this block is only executed if character is
# in the string
self.empty[index] = choice # no need to pop, you can simply change the value of the list at a given index
else:
print "wrong"
self.turns += 1
print self.empty
I am new to python. Trying to write a script that will use numeric colomns from a file whcih also contains a header. Here is an example of a file:
#File_Version: 4
PROJECTED_COORDINATE_SYSTEM
#File_Version____________-> 4
#Master_Project_______->
#Coordinate_type_________-> 1
#Horizon_name____________->
sb+
#Horizon_attribute_______-> STRUCTURE
474457.83994 6761013.11978
474482.83750 6761012.77069
474507.83506 6761012.42160
474532.83262 6761012.07251
474557.83018 6761011.72342
474582.82774 6761011.37433
474607.82530 6761011.02524
I'd like to skip the header. here is what i tried. It works of course if i know which characters will appear in the header like "#" and "#". But how can i skip all lines containing any letter character?
in_file1 = open(input_file1_short, 'r')
out_file1 = open(output_file1_short,"w")
lines = in_file1.readlines ()
x = []
y = []
for line in lines:
if "#" not in line and "#" not in line:
strip_line = line.strip()
replace_split = re.split(r'[ ,|;"\t]+', strip_line)
x = (replace_split[0])
y = (replace_split[1])
out_file1.write("%s\t%s\n" % (str(x),str(y)))
in_file1.close ()
Thank you very much!
I think you could use some built ins like this:
import string
for line in lines:
if any([letter in line for letter in string.ascii_letters]):
print "there is an ascii letter somewhere in this line"
This is only looking for ascii letters, however.
you could also:
import unicodedata
for line in lines:
if any([unicodedata.category(unicode(letter)).startswith('L') for letter in line]):
print "there is a unicode letter somewhere in this line"
but only if I understand my unicode categories correctly....
Even cleaner (using suggestions from other answers. This works for both unicode lines and strings):
for line in lines:
if any([letter.isalpha() for letter in line]):
print "there is a letter somewhere in this line"
But, interestingly, if you do:
In [57]: u'\u2161'.isdecimal()
Out[57]: False
In [58]: u'\u2161'.isdigit()
Out[58]: False
In [59]: u'\u2161'.isalpha()
Out[59]: False
The unicode for the roman numeral "Two" is none of those,
but unicodedata.category(u'\u2161') does return 'Nl' indicating a numeric (and u'\u2161'.isnumeric() is True).
This will check the first character in each line and skip all lines that doesn't start with a digit:
for line in lines:
if line[0].isdigit():
# we've got a line starting with a digit
Use a generator pipeline to filter your input stream.
This takes the lines from your original input lines, but stops to check that there are no letters in the entire line.
input_stream = (line in lines if
reduce((lambda x, y: (not y.isalpha()) and x), line, True))
for line in input_stream:
strip_line = ...
a_lst = ['chair','gum','food','pizza']
letter = 'x'
for word in a_lst:
if letter not in word:
print ('no',letter)
elif letter in word:
print ('yes',letter)
Output:
no x
no x
no x
no x
Is there a way i can iterate though each item in "a_lst", check if each item has letter 'x'. if no item has letter 'x' print 'no x' just Once. If a word contains letter 'x', print 'yes x' just Once.
I think my logic is flawed somewhere.
Any suggestions?
Thanks
You could do this:
letter = 'x'
result = [a for a in a_lst if letter in a]
if result:
print('yes', letter)
else:
print('no', letter)
Explanation:
result will be [] if none of the words in a_lst has the letter. When you do a if result on an empty list, it returns False, otherwise it returns True. The conditional statements check and print the output statement accordingly.
Another way to do it in python is to use the filter function:
if filter(lambda x: letter in x, a_lst):
print('yes', letter)
else:
print('no', letter)
Yet another way to do it is to use any:
if any(letter in word for word in a_list):
print('yes', letter)
else:
print('no', letter)
any(letter in word for word in a_list) returns True if any of the words have the letter.
You can use the any function!
if any(letter in word for word in a_lst):
print('yes', letter)
else:
print('no', letter)
Have you tried something like this:
a = ['chair', 'gum', 'food', 'pizza']
letter = 'a'
result = 'no ' + letter
k = 0
for i in a:
if letter in a[k]:
print(a[k])
result = 'yes ' + letter
k += 1
print(result)