How high is inflation? And is it the same in every country?

The latest Bamboo Weekly poses six inflation-related questions to solve with #Python #Pandas, using the latest OECD data.

Read it at: https://BambooWeekly.com

Retrieve selected items from a #Python #Pandas series by broadcasting a condition, then using it as a boolean ("mask") index:

s = pd.Series([10, 20, 30], index=list('abc'))
s >= 20 # [False, True, True]
s.loc[ s >= 20 ] # returns [20, 30]

Retrieve from a #Python #Pandas series with .loc and indexes:

s = pd.Series([10, 20, 30], index=list('abc'))
s.loc[['a', 'c']] # [10, 30]

If you pass booleans, True means "return a value," and False means "ignore it."

s.loc[[True, False, True]] # [10, 30]

#travel #travelguides #pandas #southkorea

https://www.koreatimes.co.kr/business/companies/20260617/everlands-famous-twin-pandas-become-koreas-newest-travel-guides?utm_source=flipboard&utm_medium=activitypub

Posted into THE TRAVEL EXCHANGE 3.0 @the-travel-exchange-3-0-Econopass

Apply an operator to a #Python #Pandas series, with a scalar value, *broadcasts*:

s = pd.Series([10, 20, 30], index=list('abc'))

Arithmetic is most obvious:

s + 5 # [15, 25, 35]
s ** 2 # [100, 400, 900]

But comparisons work, too:

s >= 20 # [False, True, True]

What's the fastest way to retrieve the first 2 items from a #Python #Pandas series?

s = pd.Series([10, 20, 30], index=list('abc'))

s.loc[['a','b']] # 108 µs
s.loc['a':'b'] # 22.7 µs
s.iloc[[0, 1]] # 26.7 µs
s.iloc[:2] # 18 µs
s.head(2) # 22.3 µs

Using .loc to retrieve from a #Python #Pandas series is more convenient. But .iloc is faster:

s = pd.Series([10, 20, 30], index=list('abc'))

%timeit s.loc['a'] # 2.09 μs
%timeit s.iloc[0] # 1.5 μs

%timeit s.loc[['a', 'b']] # 105 μs
%timeit s.iloc[[0, 1]] # 26.4 μs

Retrieve from a #Python #Pandas series by position, rather than the index, with iloc:

s = pd.Series([10, 20, 30], index=list('abc'))

s.loc['a'] # 10
s.loc[['a', 'b']] # series of a:10, b:20

s.iloc[0] # 10
s.iloc[[0, 1]] # series of a:10, b:20

Pass a list to .loc on a #Python #Pandas series, and get a series back:

s = pd.Series([10, 20, 30], index=list('abc'))

s.loc[['a']] # series, a:10
s.loc[['a', 'b']] # series, a:10, b:20
s.loc[['a', 'a']] # series, a:10 twice!
s.loc[['b', 'a']] # series, b:20, a:10