Background
The following question is a variation from Unnest grab keywords/nextwords/beforewords function.
1) I have the following word_list
word_list = ['crayons', 'cars', 'camels']
2) And df1
l = ['there are many crayons, in the blue box crayons that are',
'cars! i like a lot of sports cars because they go fast',
'the camels, in the middle east have many camels to ride ']
df1 = pd.DataFrame(l, columns=['Text'])
df1
Text
0 there are many crayons, in the blue box crayons that are
1 cars! i like a lot of sports cars because they go fast
2 the camels, in the middle east have many camels to ride
3) I also have a function find_next_words which uses word_list to grab words from Text column in df1
def find_next_words(row, word_list):
sentence = row[0]
trigger_words = []
next_words = []
for keyword in word_list:
words = sentence.split()
for index in range(0, len(words) - 1):
if words[index] == keyword:
trigger_words.append(keyword)
next_words.append(words[index + 1:index + 3])
return pd.Series([trigger_words, next_words], index = ['TriggerWords','NextWords'])
4) And it's pieced together with the following
df2 = df1.join(df.apply(lambda x: find_next_words(x, word_list), axis=1))
Output
Text TriggerWords NextWords
0 [crayons] [[that, are]]
1 [cars] [[because, they]]
2 [camels] [[to, ride]]
Problem
5) The output misses the following
crayons, from row 0 of Text column df1
cars! from row 1 of Text column df1
camels, from row 2 of Text column df1
Goal
6) Grab all corresponding words from df1 even if the words in df1 have a slight variation e.g. crayons, cars! from the words in word_list
(For this toy example, I know I can easily fix this problem by just adding these word variations to word_list = ['crayons,','crayons', 'cars!',cars, 'camels,', 'camels']. But this would be impractical to do with my my real word_list, which contains ~20K words)
Desired Output
Text TriggerWords NextWords
0 [crayons, crayons] [[in, the], [that, are]]
1 [cars, cars] [[i,like],[because, they]]
2 [camels, camels] [[in, the], [to, ride]]
Questions
How do I 1) tweak my word_list (e.g. regex?) 2) or find_next_words function to achieve my desired output?
You can tweak your regex something like this
\b(crayons|cars|camels)\b(?:[^a-z\n]*([a-z]*)[^a-z\n]*([a-z]*))
Regex Demo
import nltk
change
words = sentence.split()
to
words = nltk.word_tokenize(sentence)
this leads to
'crayons', ','
instead of
'crayons,'
which allows find_next_words to correctly identify all words from word_list in Text column
Related
From the following data frame:
d = {'col1':['a-1524112-124', 'b-1515', 'c-584854', 'a-15154']}
df = pd.DataFrame.from_dict(d)
My ultimate goal is to extract the letters a, b or c (as string) in a pandas series. For that I am using the .findall() method from the re module, as shown below:
# import the module
import re
# define the patterns
pat = 'a|b|c'
# extract the patterns from the elements in the specified column
df['col1'].str.findall(pat)
The problem is that the output i.e. the letters a, b or c, in each row, will be present in a list (of a single element), as shown below:
Out[301]:
0 [a]
1 [b]
2 [c]
3 [a]
While I would like to have the letters a, b or c as string, as shown below:
0 a
1 b
2 c
3 a
I know that if I combine re.search() with .group() I can get a string, but if I do:
df['col1'].str.search(pat).group()
I will get the following error message:
AttributeError: 'StringMethods' object has no attribute 'search'
Using .str.split() won't do the job because, in my original dataframe, I want to capture strings that might contain the delimiter (e.g. I might want to capture a-b)
Does anyone know a simple solution for that, perhaps avoiding iterative operations such as a for loop or list comprehension?
Use extract with capturing groups:
import pandas as pd
d = {'col1':['a-1524112-124', 'b-1515', 'c-584854', 'a-15154']}
df = pd.DataFrame.from_dict(d)
result = df['col1'].str.extract('(a|b|c)')
print(result)
Output
0
0 a
1 b
2 c
3 a
Fix your code
pat = 'a|b|c'
df['col1'].str.findall(pat).str[0]
Out[309]:
0 a
1 b
2 c
3 a
Name: col1, dtype: object
Simply try with str.split() like this- df["col1"].str.split("-", n = 1, expand = True)
import pandas as pd
d = {'col1':['a-1524112-124', 'b-1515', 'c-584854', 'a-15154']}
df = pd.DataFrame.from_dict(d)
df['col1'] = df["col1"].str.split("-", n = 1, expand = True)
print(df.head())
Output:
col1
0 a
1 b
2 c
3 a
I have a DataFrame in which one column contains lists of words.
>>dataset.head(1)
>> contain
0 ["name", "Place", "ect#gtr", "nick"]
1 ["gf#e", "nobel", "play", "hi"]
I want to remove all the words which contain '#'. In the above example, I want to remove "ect#gtr" and "gf#e".
Try This one
ab= np.column_stack([~df[col].str.contains(r"#") for col in df])
new_df=df.loc[ab.any(axis=1)]
print(new_df)
Use list comprehension with filtering, regex here is not necessary:
df = pd.DataFrame({'contain':[['name', 'Place', 'ect#gtr', 'nick'],
['gf#e', 'nobel', 'play', 'hi']]})
print (df)
contain
0 [name, Place, ect#gtr, nick]
1 [gf#e, nobel, play, hi]
df.contain = df.contain.apply(lambda x: [y for y in x if '#' not in y])
Or:
df.contain = [[y for y in x if '#' not in y] for x in df.contain]
print (df)
contain
0 [name, Place, nick]
1 [nobel, play, hi]
EDIT: For remove values in strings add split with join:
df = pd.DataFrame({'contain':['name Place ect#gtr nick',"gf#e nobel play hi"]})
print (df)
contain
0 name Place ect#gtr nick
1 gf#e nobel play hi
df.contain = df.contain.apply(lambda x: ' '.join([y for y in x.split() if '#' not in y]))
print (df)
contain
0 name Place nick
1 nobel play hi
I have three lists that look like this:
age = ['51+', '21-30', '41-50', '31-40', '<21']
cluster = ['notarget', 'cluster3', 'allclusters', 'cluster1', 'cluster2']
device = ['htc_one_2gb','iphone_6/6+_at&t','iphone_6/6+_vzn','iphone_6/6+_all_other_devices','htc_one_2gb_limited_time_offer','nokia_lumia_v3','iphone5s','htc_one_1gb','nokia_lumia_v3_more_everything']
I also have column in a df that looks like this:
campaign_name
0 notarget_<21_nokia_lumia_v3
1 htc_one_1gb_21-30_notarget
2 41-50_htc_one_2gb_cluster3
3 <21_htc_one_2gb_limited_time_offer_notarget
4 51+_cluster3_iphone_6/6+_all_other_devices
I want to split the column into three separate columns based on the values in the above lists. Like so:
age cluster device
0 <21 notarget nokia_lumia_v3
1 21-30 notarget htc_one_1gb
2 41-50 cluster3 htc_one_2gb
3 <21 notarget htc_one_2gb_limited_time_offer
4 51+ cluster3 iphone_6/6+_all_other_devices
First thought was to do a simple test like this:
ages_list = []
for i in ages:
if i in df['campaign_name'][0]:
ages_list.append(i)
print ages_list
>>> ['<21']
I was then going to convert ages_list to a series and combine it with the remaining two to get the end result above but i assume there is a more pythonic way of doing it?
the idea behind this is that you'll create a regular expression based on the values you already have , for example if you want to build a regular expressions that capture any value from your age list you may do something like this '|'.join(age) and so on for all the values you already have cluster & device.
a special case for device list becuase it contains + sign that will conflict with the regex ( because + means one or more when it comes to regex ) so we can fix this issue by replacing any value of + with \+ , so this mean I want to capture literally +
df = pd.DataFrame({'campaign_name' : ['notarget_<21_nokia_lumia_v3' , 'htc_one_1gb_21-30_notarget' , '41-50_htc_one_2gb_cluster3' , '<21_htc_one_2gb_limited_time_offer_notarget' , '51+_cluster3_iphone_6/6+_all_other_devices'] })
def split_df(df):
campaign_name = df['campaign_name']
df['age'] = re.findall('|'.join(age) , campaign_name)[0]
df['cluster'] = re.findall('|'.join(cluster) , campaign_name)[0]
df['device'] = re.findall('|'.join([x.replace('+' , '\+') for x in device ]) , campaign_name)[0]
return df
df.apply(split_df, axis = 1 )
if you want to drop the original column you can do this
df.apply(split_df, axis = 1 ).drop( 'campaign_name', axis = 1)
Here I'm assuming that a value must be matched by regex but if this is not the case you can do your checks , you got the idea
I already loaded 20 csv files with function:
tbl = list.files(pattern="*.csv")
list_of_data = lapply(tbl, read.csv)
I combined all of those filves into one:
all_data = do.call(rbind.fill, list_of_data)
In the new table is a column called "Accession". After combining many of the names (Accession) are repeated. And I would like to remove all of the duplicates.
Another problem is that some of those "names" are ALMOST the same. The difference is that there is name and after become the dot and the number.
Let me show you how it looks:
AT3G26450.1 <--
AT5G44520.2
AT4G24770.1
AT2G37220.2
AT3G02520.1
AT5G05270.1
AT1G32060.1
AT3G52380.1
AT2G43910.2
AT2G19760.1
AT3G26450.2 <--
<-- = Same sample, different names. Should be treated as one. So just ignore dot and a number after.
Tried this one:
all_data$CleanedAccession = str_extract(all_data$Accession, "^[[:alnum:]]+")
all_data = subset(all_data, !duplicated(CleanedAccession))
Error in `$<-.data.frame`(`*tmp*`, "CleanedAccession", value = character(0)) :
You can use this command to both subset and rename the values:
subset(transform(alldata, Ascension = sub("\\..*", "", Ascension)),
!duplicated(Ascension))
Ascension
1 AT3G26450
2 AT5G44520
3 AT4G24770
4 AT2G37220
5 AT3G02520
6 AT5G05270
7 AT1G32060
8 AT3G52380
9 AT2G43910
10 AT2G19760
What about
df <- data.frame( Accession = c("AT3G26450.1",
"AT5G44520.2",
"AT4G24770.1",
"AT2G37220.2",
"AT3G02520.1",
"AT5G05270.1",
"AT1G32060.1",
"AT3G52380.1",
"AT2G43910.2",
"AT2G19760.1",
"AT3G26450.2"))
df[!duplicated(unlist(lapply(strsplit(as.character(df$Accession),
".", fixed = T), "[", 1))), ]
I have written a code to take a text file as input and print only the variants which repeat more than once. By variants I mean, chr positions in the text file.
The input file looks like this:
chr1 1048989 1048989 A G intronic C1orf159 0.16 rs4970406
chr1 1049083 1049083 C A intronic C1orf159 0.13 rs4970407
chr1 1049083 1049083 C A intronic C1orf159 0.13 rs4970407
chr1 1113121 1113121 G A intronic TTLL10 0.13 rs12092254
As you can see, rows 2 and 3 repeat. I'm just taking the first 3 columns and seeing if they are the same. Here, chr1 1049083 1049383 repeat in both row2 and row3. So I print out saying that there is one duplicate and it's position.
I have written the code below. Though it's doing what I want, it's quite slow. It takes me about 5 min to run on a file which have 700,000 rows. I wanted to know if there is a way to speed things up.
Thanks!
#!/usr/bin/env python
""" takes in a input file and
prints out only the variants that occur more than once """
import shlex
import collections
rows = open('variants.txt', 'r').read().split("\n")
# removing the header and storing it in a new variable
header = rows.pop()
indices = []
for row in rows:
var = shlex.split(row)
indices.append("_".join(var[0:3]))
dup_list = []
ind_tuple = collections.Counter(indices).items()
for x, y in ind_tuple:
if y>1:
dup_list.append(x)
print dup_list
print len(dup_list)
Note: In this case the entire row2 is a duplicate of row3. But this is not necessarily the case all the time. Duplicate of chr positions (first three columns) is what I'm looking for.
EDIT:
Edited the code as per the suggestion of damienfrancois. Below is my new code:
f = open('variants.txt', 'r')
indices = {}
for line in f:
row = line.rstrip()
var = shlex.split(row)
index = "_".join(var[0:3])
if indices.has_key(index):
indices[index] = indices[index] + 1
else:
indices[index] = 1
dup_pos = 0
for key, value in indices.items():
if value > 1:
dup_pos = dup_pos + 1
print dup_pos
I used, time to see how long both the code takes.
My original code:
time run remove_dup.py
14428
CPU times: user 181.75 s, sys: 2.46 s,total: 184.20 s
Wall time: 209.31 s
Code after modification:
time run remove_dup2.py
14428
CPU times: user 177.99 s, sys: 2.17 s, total: 180.16 s
Wall time: 222.76 s
I don't see any significant improvement in the time.
Some suggestions:
do not read the whole file at once ; read line by line and process it on the fly ; you'll save memory operations
let indices be a default dict and increment the value at key "_".join(var[0:3]) ; this saves the costly (guessing here, should use a profiler) collections.Counter(indices).items() step
try pypy or a python compiler
split your data in as many subsets as your computer has cores, apply the program to each subset in parallel then merge the results
HTH
A big time sink is probably the if..has_key() portion of the code. In my experience, try-except is a lot faster...
f = open('variants.txt', 'r')
indices = {}
for line in f:
var = line.split()
index = "_".join(var[0:3])
try:
indices[index] += 1
except KeyError:
indices[index] = 1
f.close()
dup_pos = 0
for key, value in indices.items():
if value > 1:
dup_pos = dup_pos + 1
print dup_pos
Another option there would be replace the four try except lines with:
indices[index] = 1 + indices.get(index,0)
This approach only tells how many lines of the lines are duplicated, and not how many times they are repeated. (So if one line is duped 3x, then it will say one...)
If you are only trying to count the duplicates and not delete or note them, you could tally the lines of the file as you go, and compare this to the length of the indices dictionary, and the difference is the number of dupe lines (instead of looping back through and re-counting). This might save a little time, but gives a different answer:
#!/usr/bin/env python
f = open('variants.txt', 'r')
indices = {}
total_len=0
for line in f:
total_len +=1
var = line.split()
index = "_".join(var[0:3])
indices[index] = 1 + indices.get(index,0)
f.close()
print "Number of duplicated lines:", total_len - len(indices.keys())
I'd be curious to hear what your benchmarks are for code that does not include the has_key() test...