so I created students datafram from this list
example_scores=[('Ann', 92),('Bob',55) ]
scores_df = spark.createDataFrame(example_scores,schema=['Name','Score'])
scores_df.show()
I want to replace students score with a number.
for example if their score is between 51,60 when these dataframe will i want it to show
--Bob, 6-- and etc.
I want to use if statement but I dont know how to filter so much with in dataframe.
I tried regexp_replace, translate, but its not working.
You can write a when expression to create a new column
from pyspark.sql.functions import col, when
example_scores=[('Ann', 92),('Bob',55) ]
scores_df = spark.createDataFrame(example_scores,schema=['Name','Score'])
result_df = scores_df.withColumn("Grade", F.when((F.col("Score")>=51) & (F.col("Score")<=60),"6").otherwise("1")).select("Name","Grade")
result_df.show()
Related
I am working with a dataframe and one of the columns has for values a list of strings in each row. The list contains a number of links (each list can have a different number of links). I want to create a new column that will be based on this column of lists but keep only the links that have the keyword "uploads".
To my example, the first entry of the column is like that:
['https://seekingalpha.com/instablog/5006891-hfir/4960045-natural-gas-daily',
'https://seekingalpha.com/article/4116929-weekly-natural-gas-storage-report',
'https://static.seekingalpha.com/uploads/2017/10/26/5006891-15090647719993095_origin.png',
'https://static.seekingalpha.com/uploads/2017/10/26/5006891-15090647854075453_origin.png',
'https://static.seekingalpha.com/uploads/2017/10/26/5006891-1509065004154725_origin.png',
'https://seekingalpha.com/account/research/subscribe?slug=hfir-energy&sasource=upsell']
And I want to keep only
['https://static.seekingalpha.com/uploads/2017/10/26/5006891-15090647719993095_origin.png',
'https://static.seekingalpha.com/uploads/2017/10/26/5006891-15090647854075453_origin.png',
'https://static.seekingalpha.com/uploads/2017/10/26/5006891-1509065004154725_origin.png']
And put the clean version in a new column of the same dataframe.
Can you please suggest a way to do it?
I just found a way where I create a function that looks within a list for a specific pattern (in my case the keyword "uploads")
def clean_alt_list(list_):
list_ = [s for s in list_ if "uploads" in s]
return list_
And then I apply this function into the column I am interested in
df['clean_links'] = df['links'].apply(clean_alt_list)
IIUC, this should work for you:
df = pd.DataFrame({'url': [['https://seekingalpha.com/instablog/5006891-hfir/4960045-natural-gas-daily', 'https://seekingalpha.com/article/4116929-weekly-natural-gas-storage-report', 'https://static.seekingalpha.com/uploads/2017/10/26/5006891-15090647719993095_origin.png', 'https://static.seekingalpha.com/uploads/2017/10/26/5006891-15090647854075453_origin.png', 'https://static.seekingalpha.com/uploads/2017/10/26/5006891-1509065004154725_origin.png', 'https://seekingalpha.com/account/research/subscribe?slug=hfir-energy&sasource=upsell']]})
df = df.explode('url').reset_index(drop=True)
df[df['url'].str.contains('uploads')]
Result:
url
2 https://static.seekingalpha.com/uploads/2017/1...
3 https://static.seekingalpha.com/uploads/2017/1...
4 https://static.seekingalpha.com/uploads/2017/1...
I have a dataframe with one column of unequal list which I want to spilt into multiple columns (the item value will be the column names). An example is given below
I have done through iterrows, iterating thruough the rows and examine the list from each rows. It seem workable as my dataframe has few rows. However, I wonder if there is any clean methods
I have done through additional_df = pd.DataFrame(venue_df.location.values.tolist())
However the list break down into as below
thanks fro your help
Can you try this code: built assuming venue_df.location contains the list you have shown in the cells.
venue_df['school'] = venue_df.location.apply(lambda x: ('school' in x)+0)
venue_df['office'] = venue_df.location.apply(lambda x: ('office' in x)+0)
venue_df['home'] = venue_df.location.apply(lambda x: ('home' in x)+0)
venue_df['public_area'] = venue_df.location.apply(lambda x: ('public_area' in x)+0)
Hope this helps!
First lets explode your location column, so we can get your wanted end result.
s=df['Location'].explode()
Then lets use crosstab in that series so we can get your end result
import pandas as pd
pd.crosstab(s).unstack()
I didnt test it out cause i dont know you base_df
I have a pyspark dataframe:
Now, I want to add a new column called "countryAndState", where, for example for the first row, the value would be "USA_CA". I have tried several approaches, the last one was the following:
df_2 = df.withColumn("countryAndState", '{}_{}'.format(df.country, df.state))
I have tried with "country" and "state" instead, or with simply country and state,and also using col() but nothing seems to work. Can anyone help me solve this?
You can't use Python format strings in Spark. Use concat instead:
import pyspark.sql.functions as F
df_2 = df.withColumn("countryAndState", F.concat(F.col('country'), F.lit('_'), F.col('state')))
or concat_ws, if you need to chain many columns together with a given separator:
import pyspark.sql.functions as F
df_2 = df.withColumn("countryAndState", F.concat_ws('_', F.col('country'), F.col('state')))
I have dataframe column 'review' with content like 'Food was Awesome' and I want a new column which counts the number of repetition of each word.
name The First Years Massaging Action Teether
review A favorite in our house!
rating 5
Name: 269, dtype: object
Expecting output like ['Food':1,'was':1,'Awesome':1]
I tried with for loop but its taking too long to execute
for row in range(products.shape[0]):
try:
count_vect.fit_transform([products['review_without_punctuation'][row]])
products['word_count'][row]=count_vect.vocabulary_
except:
print(row)
I would like to do it without for loop.
I found a solution for this.
I have defined a function like this-
def Vectorize(text):
try:
count_vect.fit_transform([text])
return count_vect.vocabulary_
except:
return-1
and applied above function-
from sklearn.feature_extraction.text import CountVectorizer
count_vect = CountVectorizer()
products['word_count'] = products['review_without_punctuation'].apply(Vectorize)
This solution worked and I got vocabulary in new column.
You can get the count vector for all docs like this:
cv = CountVectorizer()
count_vectors = cv.fit_transform(products['review_without_punctuation'])
To get the count vector in array format for a particular document by index, say, the 1st doc,
count_vectors[0].toarray()
The vocabulary is in
cv.vocabulary_
To get the words that make up a count vector, say, for the 1st doc, use
cv.inverse_transform(count_vectors[0])
I have a very large data frame that I want to split ALL of the columns except first two based on a comma delimiter. So I need to logically reference column names in a loop or some other way to split all the columns in one swoop.
In my testing of the split method:
I have been able to explicitly refer to ( i.e. HARD CODE) a single column name (rs145629793) as one of the required parameters and the result was 2 new columns as I wanted.
See python code below
HARDCODED COLUMN NAME --
df[['rs1','rs2']] = df.rs145629793.str.split(",", expand = True)
The problem:
It is not feasible to refer to the actual column names and repeat code.
I then replaced the actual column name rs145629793 with columns[2] in the split method parameter list.
It results in an ERROR
'str has ni str attribute'
You can index columns by position rather than name using iloc. For example, to get the third column:
df.iloc[:, 2]
Thus you can easily loop over the columns you need.
I know what you are asking, but it's still helpful to provide some input data and expected output data. I have included random input data in my code below, so you can just copy and paste this to run, and try to apply it to your dataframe:
import pandas as pd
your_dataframe=pd.DataFrame({'a':['1,2,3', '9,8,7'],
'b':['4,5,6', '6,5,4'],
'c':['7,8,9', '3,2,1']})
import copy
def split_cols(df):
dict_of_df = {}
cols=df.columns.to_list()
for col in cols:
key_name = 'df'+str(col)
dict_of_df[key_name] = copy.deepcopy(df)
var=df[col].str.split(',', expand=True).add_prefix(col)
df=pd.merge(df, var, how='left', left_index=True, right_index=True).drop(col, axis=1)
return df
split_cols(your_dataframe)
Essentially, in this solution you create a list of the columns that you want to loop through. Then you loop through that list and create new dataframes for each column where you run the split() function. Then you merge everything back together on the index. I also:
included a prefix of the column name, so the column names did not have duplicate names and could be more easily identifiable
dropped the old column that we did the split on.
Just import copy and use the split_cols() function that I have created and pass the name of your dataframe.