Apply function with pandas dataframe - POS tagger computation time - pandas

I'm very confused on the apply function for pandas. I have a big dataframe where one column is a column of strings. I'm then using a function to count part-of-speech occurrences. I'm just not sure the way of setting up my apply statement or my function.
def noun_count(row):
x = tagger(df['string'][row].split())
# array flattening and filtering out all but nouns, then summing them
return num
So basically I have a function similar to the above where I use a POS tagger on a column that outputs a single number (number of nouns). I may possibly rewrite it to output multiple numbers for different parts of speech, but I can't wrap my head around apply.
I'm pretty sure I don't really have either part arranged correctly. For instance, I can run noun_count[row] and get the correct value for any index but I can't figure out how to make it work with apply how I have it set up. Basically I don't know how to pass the row value to the function within the apply statement.
df['num_nouns'] = df.apply(noun_count(??),1)
Sorry this question is all over the place. So what can I do to get a simple result like
string num_nouns
0 'cat' 1
1 'two cats' 1
EDIT:
So I've managed to get something working by using list comprehension (someone posted an answer, but they've deleted it).
df['string'].apply(lambda row: noun_count(row),1)
which required an adjustment to my function:
def tagger_nouns(x):
list_of_lists = st.tag(x.split())
flat = [y for z in list_of_lists for y in z]
Parts_of_speech = [row[1] for row in flattened]
c = Counter(Parts_of_speech)
nouns = c['NN']+c['NNS']+c['NNP']+c['NNPS']
return nouns
I'm using the Stanford tagger, but I have a big problem with computation time, and I'm using the left 3 words model. I'm noticing that it's calling the .jar file again and again (java keeps opening and closing in the task manager) and maybe that's unavoidable, but it's really taking far too long to run. Any way I can speed it up?

I don't know what 'tagger' is but here's a simple example with a word count that ought to work more or less the same way:
f = lambda x: len(x.split())
df['num_words'] = df['string'].apply(f)
string num_words
0 'cat' 1
1 'two cats' 2

Related

How do I reverse each value in a column bit wise for a hex number?

I have a dataframe which has a column called hexa which has hex values like this. They are of dtype object.
hexa
0 00802259AA8D6204
1 00802259AA7F4504
2 00802259AA8D5A04
I would like to remove the first and last bits and reverse the values bitwise as follows:
hexa-rev
0 628DAA592280
1 457FAA592280
2 5A8DAA592280
Please help
I'll show you the complete solution up here and then explain its parts below:
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
reversed_bits = [list_of_bits[-i] for i in range(1,len(list_of_bits)+1)]
return ''.join(reversed_bits)
df['hexa-rev'] = df['hexa'].apply(lambda x: reverse_bits(x))
There are possibly a couple ways of doing it, but this way should solve your problem. The general strategy will be defining a function and then using the apply() method to apply it to all values in the column. It should look something like this:
df['hexa-rev'] = df['hexa'].apply(lambda x: reverse_bits(x))
Now we need to define the function we're going to apply to it. Breaking it down into its parts, we strip the first and last bit by indexing. Because of how negative indexes work, this will eliminate the first and last bit, regardless of the size. Your result is a list of characters that we will join together after processing.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
The second line iterates through the list of characters, matches the first and second character of each bit together, and then concatenates them into a single string representing the bit.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
The second to last line returns the list you just made in reverse order. Lastly, the function returns a single string of bits.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
reversed_bits = [list_of_bits[-i] for i in range(1,len(list_of_bits)+1)]
return ''.join(reversed_bits)
I explained it in reverse order, but you want to define this function that you want applied to your column, and then use the apply() function to make it happen.

Pandas manipulation: matching data from other columns to one column, applied uniquely to all rows

I have a model that predicts 10 words for a particular course in order of likelihood, and I'd like the first 5 words of those words that appear in the course's description.
This is the format of the data:
course_name course_title course_description predicted_word_10 predicted_word_9 predicted_word_8 predicted_word_7 predicted_word_6 predicted_word_5 predicted_word_4 predicted_word_3 predicted_word_2 predicted_word_1
Xmath 32 Precalculus Polynomial and rational functions, exponential... directed scholars approach build african different visual cultures placed global
Xphilos 2 Morality Introduction to ethical and political philosop... make presentation weekly european ways general range questions liberal speakers
My idea is for each row to start iterating from predicted_word_1 until I get the first 5 that are in the description. I'd like to save those words in the order they appear into additional columns description_word_1 ... description_word_5. (If there are <5 predicted words in the description I plan to return NAN in the corresponding columns).
To clarify with an example: if the course_description of a course is 'Polynomial and rational functions, exponential and logarithmic functions, trigonometry and trigonometric functions. Complex numbers, fundamental theorem of algebra, mathematical induction, binomial theorem, series, and sequences. ' and its first few predicted words are irrelevantword1, induction, exponential, logarithmic, irrelevantword2, polynomial, algebra...
I would want to return induction, exponential, logarithmic, polynomial, algebra for that in that order and do the same for the rest of the courses.
My attempt was to define an apply function that will take in a row and iterate from the first predicted word until it finds the first 5 that are in the description, but the part I am unable to figure out is how to create these additional columns that have the correct words for each course. This code will currently only keep the words for one course for all the rows.
def find_top_description_words(row):
print(row['course_title'])
description_words_index=1
for i in range(num_words_per_course):
description = row.loc['course_description']
word_i = row.loc['predicted_word_' + str(i+1)]
if (word_i in description) & (description_words_index <=5) :
print(description_words_index)
row['description_word_' + str(description_words_index)] = word_i
description_words_index += 1
df.apply(find_top_description_words,axis=1)
The end goal of this data manipulation is to keep the top 10 predicted words from the model and the top 5 predicted words in the description so the dataframe would look like:
course_name course_title course_description top_description_word_1 ... top_description_word_5 predicted_word_1 ... predicted_word_10
Any pointers would be appreciated. Thank you!
If I understand correctly:
Create new DataFrame with just 100 predicted words:
pred_words_lists = df.apply(lambda x: list(x[3:].dropna())[::-1], axis = 1)
Please note that, there are lists in each row with predicted words. The order is nice, I mean the first, not empty, predicted word is on the first place, the second on the second place and so on.
Now let's create a new DataFrame:
pred_words_df = pd.DataFrame(pred_words_lists.tolist())
pred_words_df.columns = df.columns[:2:-1]
And The final DataFrame:
final_df = df[['course_name', 'course_title', 'course_description']].join(pred_words_df.iloc[:,0:11])
Hope this works.
EDIT
def common_elements(xx, yy):
temp = pd.Series(range(0, len(xx)), index= xx)
return list(df.reindex(yy).sort_values()[0:10].dropna().index)
pred_words_lists = df.apply(lambda x: common_elements(x[2].replace(',','').split(), list(x[3:].dropna())), axis = 1)
Does it satisfy your requirements?
Adapted solution (OP):
def get_sorted_descriptions_words(course_description, predicted_words, k):
description_words = course_description.replace(',','').split()
predicted_words_list = list(predicted_words)
predicted_words = pd.Series(range(0, len(predicted_words_list)), index=predicted_words_list)
predicted_words = predicted_words[~predicted_words.index.duplicated()]
ordered_description = predicted_words.reindex(description_words).dropna().sort_values()
ordered_description_list = pd.Series(ordered_description.index).unique()[:k]
return ordered_description_list
df.apply(lambda x: get_sorted_descriptions_words(x['course_description'], x.filter(regex=r'predicted_word_.*'), k), axis=1)

Creating a function to count the number of pos in a pandas instance

I've used NLTK to pos_tag sentences in a pandas dataframe from an old Yelp competition. This returns a list of tuples (word, POS). I'd like to count the number of parts of speech for each instance. How would I, say, create a function to count the number of being verbs in each review? I know how to apply functions to features - no problem there. I just can't wrap my head around how to count things inside tuples inside lists inside a pd feature.
The head is here, as a tsv: https://pastebin.com/FnnBq9rf
Thank you #zhangyulin for your help. After two days, I learned some incredibly important things (as a novice programmer!). Here's the solution!
def NounCounter(x):
nouns = []
for (word, pos) in x:
if pos.startswith("NN"):
nouns.append(word)
return nouns
df["nouns"] = df["pos_tag"].apply(NounCounter)
df["noun_count"] = df["nouns"].str.len()
As an example, for dataframe df, noun count of the column "reviews" can be saved to a new column "noun_count" using this code.
def NounCount(x):
nounCount = sum(1 for word, pos in pos_tag(word_tokenize(x)) if pos.startswith('NN'))
return nounCount
df["noun_count"] = df["reviews"].apply(NounCount)
df.to_csv('./dataset.csv')
There are a number of ways you can do that and one very straight forward way is to map the list (or pandas series) of tuples to indicator of whether the word is a verb, and count the number of 1's you have.
Assume you have something like this (please correct me if it's not, as you didn't provide an example):
a = pd.Series([("run", "verb"), ("apple", "noun"), ("play", "verb")])
You can do something like this to map the Series and sum the count:
a.map(lambda x: 1 if x[1]== "verb" else 0).sum()
This will return you 2.
I grabbed a sentence from the link you shared:
text = nltk.word_tokenize("My wife took me here on my birthday for breakfast and it was excellent.")
tag = nltk.pos_tag(text)
a = pd.Series(tag)
a.map(lambda x: 1 if x[1]== "VBD" else 0).sum()
# this returns 2

Finding the count of a set of substrings in pandas dataframe

I am given a set of substrings. I need to find the count of occurrence of all those substrings in a particular column in a dataframe. The relevant datframe would look like this
training['concat']
0 svAxu$paxArWAn
1 xvAxaSa$varRANi
2 AxAna$xurbale
3 go$BakwAH
4 viXi$Bexena
5 nIwi$kuSalaM
6 lafkA$upamam
7 yaSas$lipsoH
8 kaSa$AGAwam
9 hewumaw$uwwaram
10 varRa$pUgAn
My set of substrings is a dictionary, where the keys are the substrings and values are the probabilities with which they occur
reg = {'anuBavAn':0.35, 'a$piwra':0.2 ...... 'piwra':0.7, 'pa':0.03, 'a':0.0005}
#The length of dicitioanry is 2000
Particularly I need to find those substrings which occur more than twice
I have written the following code that performs the task. Is there a more elegant pythonic way or panda specific way to achieve the same as the current implementation is taking quite some time to execute.
elites = dict()
for reg_pat in reg_:
count = 0
eliter = len(training[training['concat'].str.contains(reg_pat)]['concat'])
if eliter >=3:
elites[reg_pat] = reg_[reg_pat]
You can use apply instead str.contains, it is faster:
reg_ = {'anuBavAn':0.35, 'a$piwra':0.2, 'piwra':0.7, 'pa':0.03, 'a':0.0005}
elites = dict()
for reg_pat in reg_:
if training['concat'].apply(lambda x: reg_pat in x).sum() >= 3:
elites[reg_pat] = reg_[reg_pat]
print (elites)
{'a': 0.0005}
Hopefully I have interpreted your question correctly. I'm inclined to stay away from regex here (in fact, I've never used it in conjunction with pandas), but it's not wrong, strictly speaking. In any case, I find it hard to believe that any regex operations are faster than a simple in check, but I could be wrong on that.
for substr in reg:
totalStringAppearances = training.apply((lambda string: substr in string))
totalStringAppearances = totalStringAppearances.sum()
if totalStringAppearances > 2:
reg[substr] = totalStringAppearances / len(training)
else:
# do what you want to with the very rare substrings
Some gotchas:
If you wanted something like a substring 'a' in 'abcdefa' to return 2, then this will not work. It merely checks for existence of the substring in each string.
Inside the apply(), I am using a potentially unreliable exploitation of booleans. See this question for more details.
Post-edit: Jezrael's answer is more complete as it uses the same variable names. But, in a simple case, regarding regex vs. apply and in, I validate his claim, and my presumption:

Dynamically creating variables, while doing map/apply on a dataframe in pandas to get key names for the values in Series object returned

I am writing code for a Naive Bayes model(I know there's a standard implementation in Sklearn, but I want to code it anyway) - For this I have say upwards of 30 features, against all of which I have the corresponding click & impression counts (Treat them as True/False flags)
What I need then, is to calculate
P(Click/F1, F2.. F30) = (P(Click)*P(F1/Click)*P(F2|click) ..*P(F30|Click))/(P(F1, F2...F30), and
P(NoClick/F1, F2.. F30) = (P(NoClick)*P(F1/NoClick)*P(F2|Noclick) ..*P(F30|NOClick))/(P(F1, F2...F30)
Where I will disregard the denominator as it will affect both Click & Non click behaviour similarly.
Example, for two features, day_custom & is_tablet_phone, I have
is_tablet_phone click impression
FALSE 375417 28291280
TRUE 17743 4220980
day_custom click impression
Fri 77592 7029703
Mon 43576 3773571
Sat 65950 5447976
Sun 66460 5031271
Thu 74329 6971541
Tue 55282 4575114
Wed 51555 4737712
My approach to the Problem : Assuming I read the individual files in data frame, one after another, I want the abilty to calculate & store the corresponding Probablities back in a file, that I will then use for real time prediction of Probabilty to click vs no click.
One possible structure of "processed file" thus would be -:
Here's my entire code -:
In the full blown example, I am traversing the entire directory structure(of 30 txt files, one at a time, from the base path) - which is why I need the ability to create "names" at runtime.
for base_path in base_paths:
for root, dirs, files in os.walk(base_path):
for file in files:
file_paths.append(os.path.join(root, file))
For reasons of tractability, follow from here, by taking the 2 txt files as sample input
file_paths=['/home/ekta/Desktop/NB/day_custom.txt','/home/ekta/Desktop/NB/is_tablet_phone.txt']
flag=0
for filehandle in file_paths:
feature_name=filehandle.split("/")[-1].split(".")[0]
df= pd.read_csv(filehandle,skiprows=0, encoding='utf-8',sep='\t',index_col=False,dtype={feature_name: object,'click': int,'impression': int})
df2=df[(df.impression-df.click>0) & (df.click >0)]
if flag ==0:
MySumC,MySumNC,Mydict=0,0,collections.defaultdict(dict)
MySumC=sum(df2['click'])
MySumNC=sum(df2['impression'])
P_C=float(MySumC)/float(MySumC+MySumNC)
P_NC=1-P_C
for feature_value in df2[feature_name]:
Mydict[feature_name+'_'+feature_value]={'P_'+feature_name+'_'+feature_value+'_C':(df2[df2[feature_name]==feature_value]['click']*float(P_C))/MySumC, \
'P_'+feature_name+'_'+feature_value+'_NC':(df2[df2[feature_name]==feature_value]['impression']*float(P_NC))/MySumNC}
flag=1 %Set the flag as "1" because we don't need to compute the MySumC,MySumNC, P_C & P_NC again
Question :
It looks like THIS loop is the killer here.Also, intutively, looping on a dataframe is a BAD practice. How can I rewrite this, perhaps using Map/Apply ?
for feature_value in df2[feature_name]:
Mydict[feature_name+'_'+feature_value]={'P_'+feature_name+'_'+feature_value+'_C':(df2[df2[feature_name]==feature_value]['click']*float(P_C))/MySumC, \
'P_'+feature_name+'_'+feature_value+'_NC':(df2[df2[feature_name]==feature_value]['impression']*float(P_NC))/MySumNC}
What I need in Mydict , which is a hash to store each feature name and each feature value in it
{'day_custom_Mon':{'P_day_custom_Mon_C':.787,'P_day_custom_Mon_NC': 0.556},
'day_custom_Tue':{'P_day_custom_Tue_C':0.887,'P_day_custom_Tue_NC': 0.156},
'day_custom_Wed':{'P_day_custom_Tue_C':0.087,'P_day_custom_Tue_NC': 0.167}
'day_custom_Thu':{'P_day_custom_Tue_C':0.947,'P_day_custom_Tue_NC': 0.196},
'is_tablet_phone_True':{'P_is_tablet_phone_True_C':.787,'P_is_tablet_phone_True_NC': 0.066},
'is_tablet_phone_False':{'P_is_tablet_phone_False_C':.787,'P_is_tablet_phone_False_NC': 0.077},
.. and so on..
%PPS: I just made up those float numbers, but you get the point
Also because I will later serialize this file & pass to Redis directly, for other systems to feed on it, in an cron-job manner, so I need to preserve some sort of Dynamic naming .
What I tried -:
Since I am reading feature_name as
feature_name=filehandle.split("/")[-1].split(".")[0]` # thereby abstracting & creating variables dynamically
def funct1(row):
return row[feature_name]
def funct2(row):
return row['click']
def funct3(row):
return row['impression']
then..
df2.apply(funct2,axis=1)df2.apply(funct,axis=1)*float(P_C))/MySumC, df2.apply(funct3,axis=1)*float(P_NC))/MySumNC Gives me both the values I need for a feature_value(say Mon, Tue, Wed, and so on..) for a feature_name (say,day_custom)
I also know that df2.apply(funct1, axis=1) contains part of mycustom "names"(ie feature values), how would I then build these names using map/apply ?
Ie. I will have the values, but how would I create the "key" 'P_'+feature_name+'_'+feature_value+'_C' , since feature value post apply is returned as a series object.
check out the following recipe which does exactly what you want, only using data frame manipulations. I also simplified the actual frequency calculation a bit ;)
#set the feature name values as the index of
df2.set_index(feature_name, inplace=True)
#This is what df2.set_index() looks like:
# click impression
#day_custom
#Fri 9917 3163
#Mon 2566 3818
#Sat 8725 7753
#Sun 6938 8642
#Thu 6136 2556
#Tue 5234 2356
#Wed 9463 9433
#rename the index of your data frame
df2.rename(index=lambda x:"%s_%s"%('day_custom', x), inplace=True)
#compute the total sum of your data frame entries
totsum = float(df2.values.sum())
#use apply to multiply every data frame element by the total sum
df2 = df2.applymap(lambda x:x/totsum)
#transpose the data frame to have the following shape
#day_custom day_custom_Fri day_custom_Mon ...
#click 0.102019 0.037468 ...
#impression 0.087661 0.045886 ...
#
#
dftranspose = df2.T
# template kw for formatting
templatekw = {'click':"P_%s_C", 'impression':"P_%s_NC"}
# build a list of small data frames with correct index names P_%s_NC etc
dflist = [dftranspose[[col]].rename(lambda x:templatekw[x]%col) for col in dftranspose]
#use the concatenate function to produce a sparse dictionary
MyDict= pd.concat(dflist).to_dict()
Instead of assigning to MyDict at the end, you can use the update-method during the loop.
For understanding the comments below, see here my
Original answer:
Try to use a pivot_table:
def clickfunc(x):
return np.sum(x) * P_C / MySumC
def impressionfunc(x):
return np.sum(x) * P_NC / MySumNC
newtable = df2.pivot_table(['click', 'impression'], 'feature_name', \
aggfunc=[clickfunc, impressionfunc])
#transpose the table for the dictionary to have the right form
newtable = newtable.T
#to_dict functionality already gives the correct result
MyDict = newtable.to_dict()
#rename by copying
for feature_value, subdict in MyDict.items():
word = feature_name +"_"+ feature_value
copydict[word] = {'P_' + word + '_C':subdict['click'],\
'P_' + word + '_NC':subdict['impression'] }
This gives you the result you want in copydict
itertuples() is what worked for me(worked at lightspeed) - though It is still not using the map/apply approach that I so much wanted to see. Itertuples on a pandas dataframe returns the whole row, so I no longer have to do df2[df2[feature_name]==feature_value]['click'] - be aware that this matching by value is not only expensive, but also undesired, since it may return a series, if there were duplicate rows. itertuples solves that problem were elegantly, though I need to then access the individual objects/columns by integer indexes , which means less re-usable code. I could abstract this, but It wont be like accessing by column names, the status-quo.
for row in df2.itertuples():
Mydict[feature_name+'_'+str(row[1])]={'P_'+feature_name+'_'+str(row[1])+'_C':(row[2]*float(P_C))/MySumC, \
'P_'+feature_name+'_'+str(row[1])+'_NC':(row[3]*float(P_NC))/MySumNC}
Note that I am accesing each column in the row by row[1] , row[2] and like. For example, row has (0, u'Fri', 77592, 7029703)
Post this I get
dict(Mydict)
{'day_custom_Thu': {'P_day_custom_Thu_NC': 0.18345372640838162, 'P_day_custom_Thu_C': 0.0019559423132143377}, 'day_custom_Mon': {'P_day_custom_Mon_C': 0.0011466875948906617, 'P_day_custom_Mon_NC': 0.099300235316209587}, 'day_custom_Sat': {'P_day_custom_Sat_NC': 0.14336163246883712, 'P_day_custom_Sat_C': 0.0017354517827023852}, 'day_custom_Tue': {'P_day_custom_Tue_C': 0.001454726996987919, 'P_day_custom_Tue_NC': 0.1203925662982053}, 'day_custom_Sun': {'P_day_custom_Sun_NC': 0.13239618235343156, 'P_day_custom_Sun_C': 0.0017488722589598259}, 'is_tablet_phone_TRUE': {'P_is_tablet_phone_TRUE_NC': 0.11107365073163174, 'P_is_tablet_phone_TRUE_C': 0.00046690100046229593}, 'day_custom_Wed': {'P_day_custom_Wed_NC': 0.12467127727567069, 'P_day_custom_Wed_C': 0.0013566522616712882}, 'day_custom_Fri': {'P_day_custom_Fri_NC': 0.1849842396242351, 'P_day_custom_Fri_C': 0.0020418070466026303}, 'is_tablet_phone_FALSE': {'P_is_tablet_phone_FALSE_NC': 0.74447539516197614, 'P_is_tablet_phone_FALSE_C': 0.0098789704610580936}}