In the example I made up below, I am trying to get the words STEM Employment from the text list. If I find those set of words in order, I would like to find the index number of the first word so I can then use that same index number for the width and height lists since they are parallel (meaning their len is always the same dynamic value)
teststring = ("STEM Employment").split(" ")
data = {"text":["some","more","STEM","Employment","data"],
"width":[100,45,50,90,354],
"height":[500,320,320,432,554]}
so for this example, the answer would be 50 and 320 because the first word is STEM. However I am not just looking for STEM I have to make sure that Employment follows right after STEM in the list.
I tried writing a forloop for this but my forloop stops short when it confirms the first word STEM. I am not sure how to fix it:
testchecker = 0
for testword in range(len(data)):
print(data["text"][testword])
for m in teststring:
# print(m)
print(testchecker)
if m in data["text"][testword]:
print("true")
testchecker = testchecker + 1
if testchecker == len(teststring):
print("match")
print(testword-testchecker+1)
pass
else:
testchecker = 0
You can make data["text"] a string with join and check for "STEM Employment" in that. Then find the index of "STEM".
teststring = "STEM Employment"
data = {"text":["some","more","STEM","Employment","data"],
"width":[100,45,50,90,354],
"height":[500,320,320,432,554]}
if teststring in " ".join(data["text"]):
idx = data["text"].index(teststring.split(' ')[0])
print(data["width"][idx], data["height"][idx])
Output:
50 320
Another option:
teststring = "STEM Employment".split(' ')
# Make sure all words in testring are in data["text"]
if all(s in data["text"] for s in teststring):
# Get the indexes of each word
indexes = [data["text"].index(s) for s in teststring]
# Make sure all indexes are sequential
if all(b - a == 1 for a, b in zip(indexes, indexes[1:])):
print(data["width"][indexes[0]], data["height"][indexes[0]])
how to find Python program to find the 3rd position of occurrence of a given string in another given string.
find_string("I am the the champion of the champions of the champions", "the")
You can use a regular expression like this to find your 'needle' in a 'haystack'
import re
haystack = "I am the the champion of the champions of the champions"
needle = "the"
# print all matches
for i, match in enumerate(re.finditer(needle, haystack)):
print(f"{i+1}, start:{match.start()}, end:{match.end()}")
# or select the third from a list
matches = list(re.finditer(needle, haystack)) # list of all matches
matches[2].start() # get start position of third match
matches[2].end() # get end position of third match
EDIT:
Using only str.find
def find_nth(haystack, needle, n):
start, lneedle = haystack.find(needle), len(needle)
for _ in range(n-1):
start = haystack.find(needle, start + lneedle)
return start
find_nth(haystack, needle, 3)
You can do like this,
def find_string(txt, str1, n):
index = 0
for i in range(n + 1):
index = txt.find(str1, index + 1)
return txt.find(str1, index)
Output:
find_string("I am the the champion of the champions of the champions", "the", 3)
# Output
# 42
I am trying to attempt the mapreduce pairs pattern in python. Need to check if a word is in a text file and then find the word next to it and yield a pair of both words. keep running into either:
neighbors = words[words.index(w) + 1]
ValueError: substring not found
or
ValueError: ("the") is not in list
file cwork_trials.py
from mrjob.job import MRJob
class MRCountest(MRJob):
# Word count
def mapper(self, _, document):
# Assume document is a list of words.
#words = []
words = document.strip()
w = "the"
neighbors = words.index(w)
for word in words:
#searchword = "the"
#wor.append(str(word))
#neighbors = words[words.index(w) + 1]
yield(w,1)
def reducer(self, w, values):
yield(w,sum(values))
if __name__ == '__main__':
MRCountest.run()
Edit:
Trying to use the pairs pattern to search a document for every instance of a specific word and then find the word next to it each time. Then yielding a pair result for each instance i.e. find instances of "the" and the word next to it i.e. [the], [book], [the], [cat] etc.
from mrjob.job import MRJob
class MRCountest(MRJob):
# Word count
def mapper(self, _, document):
# Assume document is a list of words.
#words = []
words = document.split(" ")
want = "the"
for w, want in enumerate(words, 1):
if (w+1) < len(words):
neighbors = words[w + 1]
pair = (want, neighbors)
for u in neighbors:
if want is "the":
#pair = (want, neighbors)
yield(pair),1
#neighbors = words.index(w)
#for word in words:
#searchword = "the"
#wor.append(str(word))
#neighbors = words[words.index(w) + 1]
#yield(w,1)
#def reducer(self, w, values):
#yield(w,sum(values))
if __name__ == '__main__':
MRCountest.run()
As it stands I get yields of every word pair with multiples of the same pairing.
When you use words.index("the") then you will only get the first instance of "the" in your list or string, and as you have found, you will get an error if "the" isn't present.
Also you mention that you are trying to produce pairs, but only yield a single word.
I think what you are trying to do is something more like this:
def get_word_pairs(words):
for i, word in enumerate(words):
if (i+1) < len(words):
yield (word, words[i + 1]), 1
if (i-1) > 0:
yield (word, words[i - 1]), 1
assuming you are interested in neighbours in both directions. (If not, you only need the first yield.)
Lastly, since you use document.strip(), I suspect that document is in fact a string and not a list. If that's the case, you can use words = document.split(" ") to get the word list, assuming you don't have any punctuation.
I have long file like 1200 sequences
>3fm8|A|A0JLQ2
CFLVNLNADPALNELLVYYLKEHTLIGSANSQDIQLCGMGILPEHCIIDITSEGQVMLTP
QKNTRTFVNGSSVSSPIQLHHGDRILWGNNHFFRLNLP
>2ht9|A|A0JLT0
LATAPVNQIQETISDNCVVIFSKTSCSYCTMAKKLFHDMNVNYKVVELDLLEYGNQFQDA
LYKMTGERTVPRIFVNGTFIGGATDTHRLHKEGKLLPLVHQCYL
I want to read each possible pattern has cysteine in middle and has in the beginning five string and follow by other five string such as xxxxxCxxxxx
the output should be like this:
QDIQLCGMGIL
ILPEHCIIDIT
TISDNCVVIFS
FSKTSCSYCTM
this is the pogram only give position of C . it is not work like what I want
pos=[]
def find(ch,string1):
for i in range(len(string1)):
if ch == string1[i]:
pos.append(i)
return pos
z=find('C','AWERQRTCWERTYCTAAAACTTCTTT')
print z
You need to return outside the loop, you are returning on the first match so you only ever get a single character in your list:
def find(ch,string1):
pos = []
for i in range(len(string1)):
if ch == string1[i]:
pos.append(i)
return pos # outside
You can also use enumerate with a list comp in place of your range logic:
def indexes(ch, s1):
return [index for index, char in enumerate(s1)if char == ch and 5 >= index <= len(s1) - 6]
Each index in the list comp is the character index and each char is the actual character so we keep each index where char is equal to ch.
If you want the five chars that are both sides:
In [24]: s="CFLVNLNADPALNELLVYYLKEHTLIGSANSQDIQLCGMGILPEHCIIDITSEGQVMLTP QKNTRTFVNGSSVSSPIQLHHGDRILWGNNHFFRLNLP"
In [25]: inds = indexes("C",s)
In [26]: [s[i-5:i+6] for i in inds]
Out[26]: ['QDIQLCGMGIL', 'ILPEHCIIDIT']
I added checking the index as we obviously cannot get five chars before C if the index is < 5 and the same from the end.
You can do it all in a single function, yielding a slice when you find a match:
def find(ch, s):
ln = len(s)
for i, char in enumerate(s):
if ch == char and 5 <= i <= ln - 6:
yield s[i- 5:i + 6]
Where presuming the data in your question is actually two lines from yoru file like:
s="""">3fm8|A|A0JLQ2CFLVNLNADPALNELLVYYLKEHTLIGSANSQDIQLCGMGILPEHCIIDITSEGQVMLTPQKNTRTFVNGSSVSSPIQLHHGDRILWGNNHFFRLNLP
>2ht9|A|A0JLT0LATAPVNQIQETISDNCVVIFSKTSCSYCTMAKKLFHDMNVNYKVVELDLLEYGNQFQDALYKMTGERTVPRIFVNGTFIGGATDTHRLHKEGKLLPLVHQCY"""
Running:
for line in s.splitlines():
print(list(find("C" ,line)))
would output:
['0JLQ2CFLVNL', 'QDIQLCGMGIL', 'ILPEHCIIDIT']
['TISDNCVVIFS', 'FSKTSCSYCTM', 'TSCSYCTMAKK']
Which gives six matches not four as your expected output suggest so I presume you did not include all possible matches.
You can also speed up the code using str.find, starting at the last match index + 1 for each subsequent match
def find(ch, s):
ln, i = len(s) - 6, s.find(ch)
while 5 <= i <= ln:
yield s[i - 5:i + 6]
i = s.find(ch, i + 1)
Which will give the same output. Of course if the strings cannot overlap you can start looking for the next match much further in the string each time.
My solution is based on regex, and shows all possible solutions using regex and while loop. Thanks to #Smac89 for improving it by transforming it into a generator:
import re
string = """CFLVNLNADPALNELLVYYLKEHTLIGSANSQDIQLCGMGILPEHCIIDITSEGQVMLTPQKNTRTFVNGSSVSSPIQLHHGDRILWGNNHFFRLNLP
LATAPVNQIQETISDNCVVIFSKTSCSYCTMAKKLFHDMNVNYKVVELDLLEYGNQFQDA LYKMTGERTVPRIFVNGTFIGGATDTHRLHKEGKLLPLVHQCYL"""
# Generator
def find_cysteine2(string):
# Create a loop that will utilize regex multiple times
# in order to capture matches within groups
while True:
# Find a match
data = re.search(r'(\w{5}C\w{5})',string)
# If match exists, let's collect the data
if data:
# Collect the string
yield data.group(1)
# Shrink the string to not include
# the previous result
location = data.start() + 1
string = string[location:]
# If there are no matches, stop the loop
else:
break
print [x for x in find_cysteine2(string)]
# ['QDIQLCGMGIL', 'ILPEHCIIDIT', 'TISDNCVVIFS', 'FSKTSCSYCTM', 'TSCSYCTMAKK']
I'm almost done with my program, but I've made a subtle mistake. My program is supposed to take a word, and by changing one letter at a time, is eventually supposed to reach a target word, in the specified number of steps. I had been trying at first to look for similarities, for example: if the word was find, and the target word lose, here's how my program would output in 4 steps:
['find','fine','line','lone','lose]
Which is actually the output I wanted. But if you consider a tougher set of words, like Java and work, the output is supposed to be in 6 steps.
['java', 'lava', 'lave', 'wave', 'wove', 'wore', 'work']
So my mistake is that I didn't realize you could get to the target word, by using letters that don't exist in the target word or original word.
Here's my Original Code:
import string
def changeling(word,target,steps):
alpha=string.ascii_lowercase
x=word##word and target has been changed to keep the coding readable.
z=target
if steps==0 and word!= target:##if the target can't be reached, return nothing.
return []
if x==z:##if target has been reached.
return [z]
if len(word)!=len(target):##if the word and target word aren't the same length print error.
print "error"
return None
i=1
if lookup
if lookup(z[0]+x[1:]) is True and z[0]+x[1:]!=x :##check every letter that could be from z, in variations of, and check if they're in the dictionary.
word=z[0]+x[1:]
while i!=len(x):
if lookup(x[:i-1]+z[i-1]+x[i:]) and x[:i-1]+z[i-1]+x[i:]!=x:
word=x[:i-1]+z[i-1]+x[i:]
i+=1
if lookup(x[:len(x)-1]+z[len(word)-1]) and x[:len(x)-1]+z[len(x)-1]!=x :##same applies here.
word=x[:len(x)-1]+z[len(word)-1]
y = changeling(word,target,steps-1)
if y :
return [x] + y##used to concatenate the first word to the final list, and if the list goes past the amount of steps.
else:
return None
Here's my current code:
import string
def changeling(word,target,steps):
alpha=string.ascii_lowercase
x=word##word and target has been changed to keep the coding readable.
z=target
if steps==0 and word!= target:##if the target can't be reached, return nothing.
return []
if x==z:##if target has been reached.
return [z]
holderlist=[]
if len(word)!=len(target):##if the word and target word aren't the same length print error.
print "error"
return None
i=1
for items in alpha:
i=1
while i!=len(x):
if lookup(x[:i-1]+items+x[i:]) is True and x[:i-1]+items+x[i:]!=x:
word =x[:i-1]+items+x[i:]
holderlist.append(word)
i+=1
if lookup(x[:len(x)-1]+items) is True and x[:len(x)-1]+items!=x:
word=x[:len(x)-1]+items
holderlist.append(word)
y = changeling(word,target,steps-1)
if y :
return [x] + y##used to concatenate the first word to the final list, and if the/
list goes past the amount of steps.
else:
return None
The differences between the two is that the first checks every variation of find with the letters from lose. Meaning: lind, fond, fisd, and fine. Then, if it finds a working word with the lookup function, it calls changeling on that newfound word.
As opposed to my new program, which checks every variation of find with every single letter in the alphabet.
I can't seem to get this code to work. I've tested it by simply printing what the results are of find:
for items in alpha:
i=1
while i!=len(x):
print (x[:i-1]+items+x[i:])
i+=1
print (x[:len(x)-1]+items)
This gives:
aind
fand
fiad
fina
bind
fbnd
fibd
finb
cind
fcnd
ficd
finc
dind
fdnd
fidd
find
eind
fend
fied
fine
find
ffnd
fifd
finf
gind
fgnd
figd
fing
hind
fhnd
fihd
finh
iind
find
fiid
fini
jind
fjnd
fijd
finj
kind
fknd
fikd
fink
lind
flnd
fild
finl
mind
fmnd
fimd
finm
nind
fnnd
find
finn
oind
fond
fiod
fino
pind
fpnd
fipd
finp
qind
fqnd
fiqd
finq
rind
frnd
fird
finr
sind
fsnd
fisd
fins
tind
ftnd
fitd
fint
uind
fund
fiud
finu
vind
fvnd
fivd
finv
wind
fwnd
fiwd
finw
xind
fxnd
fixd
finx
yind
fynd
fiyd
finy
zind
fznd
fizd
finz
Which is perfect! Notice that each letter in the alphabet goes through my word at least once. Now, what my program does is use a helper function to determine if that word is in a dictionary that I've been given.
Consider this, instead of like my first program, I now receive multiple words that are legal, except when I do word=foundword it means I'm replacing the previous word each time. Which is why I'm trying holderlist.append(word).
I think my problem is that I need changeling to run through each word in holderlist, and I'm not sure how to do that. Although that's only speculation.
Any help would be appreciated,
Cheers.
I might be slightly confused about what you need, but by borrowing from this post I belive I have some code that should be helpful.
>>> alphabet = 'abcdefghijklmnopqrstuvwxyz'
>>> word = 'java'
>>> splits = [(word[:i], word[i:]) for i in range(len(word) + 1)]
>>> splits
[('', 'java'), ('j', 'ava'), ('ja', 'va'), ('jav', 'a'), ('java', '')]
>>> replaces = [a + c + b[1:] for a, b in splits for c in alphabet if b]
>>> replaces
['aava', 'bava', 'cava', 'dava', 'eava', 'fava', 'gava', 'hava', 'iava', 'java', 'kava', 'lava', 'mava', 'nava', 'oava', 'pava', 'qava', 'rava', 'sava', 'tava', 'uava', 'vava', 'wav
a', 'xava', 'yava', 'zava', 'java', 'jbva', 'jcva', 'jdva', 'jeva', 'jfva', 'jgva', 'jhva', 'jiva', 'jjva', 'jkva', 'jlva', 'jmva', 'jnva', 'jova', 'jpva', 'jqva', 'jrva', 'jsva', '
jtva', 'juva', 'jvva', 'jwva', 'jxva', 'jyva', 'jzva', 'jaaa', 'jaba', 'jaca', 'jada', 'jaea', 'jafa', 'jaga', 'jaha', 'jaia', 'jaja', 'jaka', 'jala', 'jama', 'jana', 'jaoa', 'japa'
, 'jaqa', 'jara', 'jasa', 'jata', 'jaua', 'java', 'jawa', 'jaxa', 'jaya', 'jaza', 'java', 'javb', 'javc', 'javd', 'jave', 'javf', 'javg', 'javh', 'javi', 'javj', 'javk', 'javl', 'ja
vm', 'javn', 'javo', 'javp', 'javq', 'javr', 'javs', 'javt', 'javu', 'javv', 'javw', 'javx', 'javy', 'javz']
Once you have a list of all possible replaces, you can simply do
valid_words = [valid for valid in replaces if lookup(valid)]
Which should give you all words that can be formed by replacing 1 character in word. By placing this code in a separate method, you could take a word, obtain possible next words from that current word, and recurse over each of those words. For example:
alphabet = 'abcdefghijklmnopqrstuvwxyz'
def next_word(word):
splits = [(word[:i], word[i:]) for i in range(len(word) + 1)]
replaces = [a + c + b[1:] for a, b in splits for c in alphabet if b]
return [valid for valid in replaces if lookup(valid)]
Is this enough help? I think your code could really benefit by separating tasks into smaller chunks.
Fixed your code:
import string
def changeling(word, target, steps):
alpha=string.ascii_lowercase
x = word #word and target has been changed to keep the coding readable.
z = target
if steps == 0 and word != target: #if the target can't be reached, return nothing.
return []
if x == z: #if target has been reached.
return [z]
holderlist = []
if len(word) != len(target): #if the word and target word aren't the same length print error.
raise BaseException("Starting word and target word not the same length: %d and %d" % (len(word),
i = 1
for items in alpha:
i=1
while i != len(x):
if lookup(x[:i-1] + items + x[i:]) is True and x[:i-1] + items + x[i:] != x:
word = x[:i-1] + items + x[i:]
holderlist.append(word)
i += 1
if lookup(x[:len(x)-1] + items) is True and x[:len(x)-1] + items != x:
word = x[:len(x)-1] + items
holderlist.append(word)
y = [changeling(pos_word, target, steps-1) for pos_word in holderlist]
if y:
return [x] + y #used to concatenate the first word to the final list, and if the list goes past the amount of steps.
else:
return None
Where len(word) and len(target), it'd be better to raise an exception than print something obscure, w/o a stack trace and non-fatal.
Oh and backslashes(\), not forward slashes(/), are used to continue lines. And they don't work on comments