Python loop through a list to find a substring match in another string from a column in dataframe then return that matched substring in a new column -


i programming in python 3.6 on spyder 3.2.3 on macos sierra 10.12.6.

i have column in dataframe df containing list of towns in australia , other information. column of interest suburbs

df["suburbs"]  apollo bay (tas.) apollo bay (vic.) apoinga act remainder - belconni

i have list states containing states in australia.

states = ["nsw", "vic", "tas", "act", "sa", "wa"]

my objective see whether suburb df["suburbs"]contains state list states , if yes, return state in column df["state"].

so, @ moment solution use loop , if statement, however, reason loop , if statement returns not found every line if matches. current loop , if statement below:

for suburb in df["suburbs"].str.upper():     state in states:         if state in suburb:             df["state"] = state         else:             df["state"] = "not found"

and returns

not found not found not found not found

another thing noticed on variable explorer section of spyder, above code creates 2 variables suburb , state values act remainder - belconni , wa, respectively. seems pick last values both dataframe column , list.

however, if not create new column state , use print function see if substring matches, shows works. code below:

for suburb in test["suburbs"].str.upper():     state in states:         if state in suburb:             print(suburb, state)

and result is:

apollo bay (tas.) tas apollo bay (vic.) vic act remainder - belconni act

it skips 1 doesn't have match. add additional else statement print not found result not found. can please me understand wrong here , why? it's pretty frustrating because seems me simple task.

thank much.


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