Replace specific words by user dictionary and others by 0

python update dictionary value list
python replace dictionary value list
replace multiple values in dictionary python
replace character in dictionary python
find and replace in dictionary python
replace list with dictionary values
python dictionary
python replace words in string from dictionary

So I have a review dataset having reviews like

Simply the best. I bought this last year. Still using. No problems faced till date.Amazing battery life. Works fine in darkness or broad daylight. Best gift for any book lover.

(This is from the original dataset, I have removed all punctuation and have all lower case in my processed dataset)

What I want to do is replace some words by 1(as per my dictionary) and others by 0. My dictionary is

dict = {"amazing":"1","super":"1","good":"1","useful":"1","nice":"1","awesome":"1","quality":"1","resolution":"1","perfect":"1","revolutionary":"1","and":"1","good":"1","purchase":"1","product":"1","impression":"1","watch":"1","quality":"1","weight":"1","stopped":"1","i":"1","easy":"1","read":"1","best":"1","better":"1","bad":"1"}

I want my output like:

0010000000000001000000000100000

I have used this code:

df['newreviews'] = df['reviews'].map(dict).fillna("0")

This always returns 0 as output. I did not want this so I took 1s and 0s as strings, but despite that I'm getting the same result. Any suggestions how to solve this?

First dont use dict as variable name, because builtins (python reserved word), then use list comprehension with get for replace not matched values to 0.

Notice:

If data are like date.Amazing - no space after punctuation is necessary replace by whitespace.

df = pd.DataFrame({'reviews':['Simply the best. I bought this last year. Still using. No problems faced till date.Amazing battery life. Works fine in darkness or broad daylight. Best gift for any book lover.']})

d = {"amazing":"1","super":"1","good":"1","useful":"1","nice":"1","awesome":"1","quality":"1","resolution":"1","perfect":"1","revolutionary":"1","and":"1","good":"1","purchase":"1","product":"1","impression":"1","watch":"1","quality":"1","weight":"1","stopped":"1","i":"1","easy":"1","read":"1","best":"1","better":"1","bad":"1"}

df['reviews']  = df['reviews'].str.replace(r'[^\w\s]+', ' ').str.lower()

df['newreviews'] = [''.join(d.get(y, '0')  for y in x.split()) for x in df['reviews']]

Alternative:

df['newreviews'] =  df['reviews'].apply(lambda x: ''.join(d.get(y, '0')  for y in x.split()))

print (df)
                                             reviews  \
0  simply the best  i bought this last year  stil...   

                        newreviews  
0  0011000000000001000000000100000  

Replace specific words by user dictionary and others by 0, replace string with dictionary python python replace dictionary value list pandas replace specific values in column replace multiple values in dictionary python Go to Home > Replace or press Ctrl+H. Enter the word or phrase you want to locate in the Find box. Enter your new text in the Replace box. Select Find Next until you come to the word you want to update.

You can do:

# clean the sentence
import re
sent = re.sub(r'\.','',sent)

# convert to list
sent = sent.lower().split()

# get values from dict using comprehension
new_sent = ''.join([str(1) if x in mydict else str(0) for x in sent])
print(new_sent)

'001100000000000000000000100000'

Replacing a specific value in a dictionary - Python, Replacing a specific value in a dictionary The user should type a letter to try guessing the hidden word. mwodict=dict(zip(mword,mwrdkey)) /// Program /// k​=0 while k<8: l = input("Please Continue until the user is out of tries or has guessed the word (in other words, letters_list has replaces star_list). Just figured out how to replace lots of different words in a txt file at one go, by iterating through a dictionary (whole word matches only). It would be really annoying if I want to replace "1" with "John", but ends up turning "12" into "John2." The following code is what works for me.

You can do it by

df.replace(repl, regex=True, inplace=True)

where df is your dataframe and repl is your dictionary.

BlackBerry Storm2 Made Simple: Written for the Storm 9500 and , 5SING�THE�3PELLING�#USTOM�$ICTIONARY� In this case, you may add these words to your own custom dictionary. you will never again be asked to replace that word with something suggested and (2) if you misspell  Show the Developer tab. If the developer tab isn't displayed in the ribbon, see Show the Developer tab.. Open a template or use a blank document. To create a form in Word that others can fill out, start with a template or document and add content controls.

A Dictionary of Chemistry and the Allied Branches of Other Sciences, that the number of times in which this replacement can take place, in other words, can form by substitution of any given acid- or alcohol-radicle for its hydrogen, CHOH fJHO(C«H«0) CH2OH CH2OH CH^PH'O) CHI0(C1H»0) Glycerin. To remove a dictionary from the list, select it and click the Remove User Dictionary icon . You must have at least one dictionary per language. If the language dictionary includes a question mark icon next to it, select the dictionary, click the Relink User Dictionary icon , and then locate and open the user dictionary.

Text, Speech and Dialogue: 5th International Conference, TSD 2002, , a file have as their marks numbers, digits can be replaced by the character #, e.g., 4.7 The Work with Words and Dictionaries Users very often need to influence work, TEA stores words – found in the testing data – with certain information. Each word is classified as, e.g., interesting, uninteresting, etc., using a scale 0, +,  It will iterate through the dictionary just like the example above and replace words (keys) found in the base_text with corresponding words (values) from the dictionary. def replace_words(base_text, device_values): for key, val in device_values.items(): base_text = base_text.replace(key, val) return base_text

The Encyclopædia Britannica, Or, Dictionary of Arts, Sciences, and , or, in other words, when one element can replace another without change of form, the We can give no further explanation of isomorphism than this, that certain (Or ammonia NII, + Arsenious acid (in its water, HIO = N II,0) unusual form). Synonyms for replace at Thesaurus.com with free online thesaurus, antonyms, and definitions. Find descriptive alternatives for replace.

Comments
  • you are not splitting the string anywhere for this map to work, also you should use dict as a variable name as it masks python's builtin dict type.
  • @AChampion how to split the string for map to work?
  • post a testable fragment of your df['reviews']
  • You probably want to do something like: df.reviews.str.split().apply(lambda review: ''.join(d.get(word, '0') for word in review)) assuming you have already lowered and removed all punctuation (and renamed dict to d).
  • Note: OP claims to already have lowered and removed punctuation, so you may be doing too much :). Also you are missing on 'Amazing' because there is no space around the punctuation - '... date.Amazing ...'
  • @AChampion - Thank you, solution should be replace punctation by whitespace.