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For the above example,
I want to have a look at the value of outlet_size when Outlet_identifier = OUT049 or any value for that instance.
I don't want to produce a new dataframe object and then print it, instead I want to know if there is any function or way to directly view it.
For both columns
df.loc[df['Outlet_identifier'].eq('OUT049'), ['Outlet_identifier', 'outlet_size']]
you can do that with pandas like this :
df = pd.DataFrame({'Outlet_identifier': ['OUT049','OUT018','OUT049'], 'outlet_size':
[2.0, 2.0, 2.0]})
df[df["Outlet_identifier"]=="OUT049"]["outlet_size"]
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import pandas as pd
Data = pd.open_csv('file name')
Data = isnull().sum()
I know that here we are testing each cell if it were empty or not; then, return the summation of empty ones.
Is there any other ways to re-write the statement?
Can we store the True results in a data structure, add them using a loop, and return the result?
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I have dataframe like this
enter image description here
I need to find out the average close days of request Recycling
Please help me.
You can group by request and get the average.This will give average for each group
df.groupby("request")["Days to Close"].mean()
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If two list is having same number, then the final list should not have the number
If your lists contains unique elements, consider using sets instead.
https://docs.python.org/2/library/sets.html
Check this code:
ls=[1,2,3,4,5,5]
ls1=[1,2,3,4,5,7,8,9]
common_elements=set(ls).intersection(set(ls1))
for i in common_elements:
if ls.__contains__(i):
ls.remove(i)
if ls1.__contains__(i):
ls1.remove(i)
final_ls=ls+ls1
print(final_ls)
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Hey guys I was wondering what exactly this code does, how does it iterate through the dataframe and what exactly does the lambda function do?
df.apply(lambda x: pd.Series(x.dropna().values))
The above block of code traverses the data frame column-wise and drops NA values. Lambda functions are the anonymous functions that are well explained in this text.
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I am using Python with Pandas. How can I multiply a column by 1000 given another column has a certain string?
This should do it.
df['columnname'] = np.where(df['othercolumn'] == 'CertainString',
df['columnname'] * 1000,
df['columnname'])