Having a df with pandas, I want to have the first index occurrence of "V1" | "V2" if possible without having to scan all the DF. Can I make it stop at the first match ?
I started doing i = df[(df["track"] == "V1") | (df["track"] == "V2")].iloc[0] but I got the full row and have a list.
Looks like you want:
df.track.isin(['V1', 'V2']).idxmax()
If you want to stop on the first match here's one way using a generator comprehension:
match = {'V1', 'V2'}
next((ix for ix, i in enumerate(df.track.values) if i in match), None)
# 1
Use a bit changed this solution - for testing use in operator - it loop only to matching like your requirement:
from numba import njit
@njit
def get_first_index_nb(A, k):
for i in range(len(A)):
if A[i] in k:
return i
return None
#pandas 0.24+
idx = get_first_index_nb(df.track.to_numpy(), ['V1', 'V2'])
#oldier pandas versions
#idx = get_first_index_nb(df.track.values, ['V1', 'V2'])
print (idx)
Solution with Series.idxmax if possible no values matching with if-else statement and Series.any, but it test all matching values:
m = df.track.isin(['V1', 'V2'])
idx = m.idxmax() if m.any() else None
Or:
idx = next(iter(df.index[df.track.isin(['V1', 'V2'])]), None)
Use Series.isin with DataFrame.index or Series.index
df[df.track.isin(['V1', 'V2'])].index[0]
or using callable
df.track.loc[lambda x: x.isin(['V1', 'V2'])].index[0]
You may use where and first_valid_index to handle case when V1 and V2 not found. In that case, first_valid_index returns None
idx = df.where(df.track.isin(['V1', 'V2'])).first_valid_index()
Just use your same code but with below changes.
From this
i = df[(df["track"] == "V1") | (df["track"] == "V2")].iloc[0]
to
df[(df["track"] == "V1") | (df["track"] == "V2")].head(1).iloc[0]
