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Removed unnecessary formatting. References to relative positions of answers are not reliable as they depend on the view (votes/oldest/active) and changing of the accepted answer and change over time (for votes, active, and accepted state)).

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edited Apr 8, 2025 at 19:58

Peter Mortensen

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In addition to all the excellent previous answers above, here is a solution to this problem when working with pandas Series objects. In many cases, pandas Series have custom/unique indices (for example, unique identifier strings) that can't be accessed with the enumerate() function.

xs = pd.Series([8, 23, 45])

xs.index = ['G923002', 'G923004', 'G923005']

print(xs)

Output:

#    G923002     8
#    G923004    23
#    G923005    45
#    dtype: int64

xs = pd.Series([8, 23, 45])

xs.index = ['G923002', 'G923004', 'G923005']

print(xs)

Output:

#    G923002     8
#    G923004    23
#    G923005    45
#    dtype: int64

We can see below that enumerate() doesn't give us the desired result:

for id, x in enumerate(xs):
    print("id #{} = {}".format(id, x))

Output:

#    id #0 = 8
#    id #1 = 23
#    id #2 = 45

for id, x in enumerate(xs):
    print("id #{} = {}".format(id, x))

Output:

#    id #0 = 8
#    id #1 = 23
#    id #2 = 45

We can access the indices of a pandas Series in a for loop using .items():

for id, x in xs.items():
    print("id #{} = {}".format(id, x))

Output:

#    id #G923002 = 8
#    id #G923004 = 23
#    id #G923005 = 45

for id, x in xs.items():
    print("id #{} = {}".format(id, x))

Output:

#    id #G923002 = 8
#    id #G923004 = 23
#    id #G923005 = 45

In addition to all the excellent answers above, here is a solution to this problem when working with pandas Series objects. In many cases, pandas Series have custom/unique indices (for example, unique identifier strings) that can't be accessed with the enumerate() function.

xs = pd.Series([8, 23, 45])

xs.index = ['G923002', 'G923004', 'G923005']

print(xs)

Output:

#    G923002     8
#    G923004    23
#    G923005    45
#    dtype: int64

We can see below that enumerate() doesn't give us the desired result:

for id, x in enumerate(xs):
    print("id #{} = {}".format(id, x))

Output:

#    id #0 = 8
#    id #1 = 23
#    id #2 = 45

We can access the indices of a pandas Series in a for loop using .items():

for id, x in xs.items():
    print("id #{} = {}".format(id, x))

Output:

#    id #G923002 = 8
#    id #G923004 = 23
#    id #G923005 = 45

In addition to all the excellent previous answers, here is a solution to this problem when working with pandas Series objects. In many cases, pandas Series have custom/unique indices (for example, unique identifier strings) that can't be accessed with the enumerate() function.

xs = pd.Series([8, 23, 45])

xs.index = ['G923002', 'G923004', 'G923005']

print(xs)

Output:

#    G923002     8
#    G923004    23
#    G923005    45
#    dtype: int64

We can see below that enumerate() doesn't give us the desired result:

for id, x in enumerate(xs):
    print("id #{} = {}".format(id, x))

Output:

#    id #0 = 8
#    id #1 = 23
#    id #2 = 45

We can access the indices of a pandas Series in a for loop using .items():

for id, x in xs.items():
    print("id #{} = {}".format(id, x))

Output:

#    id #G923002 = 8
#    id #G923004 = 23
#    id #G923005 = 45

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answered Oct 16, 2022 at 3:20

jglad

xs = pd.Series([8, 23, 45])

xs.index = ['G923002', 'G923004', 'G923005']

print(xs)

Output:

#    G923002     8
#    G923004    23
#    G923005    45
#    dtype: int64

We can see below that enumerate() doesn't give us the desired result:

for id, x in enumerate(xs):
    print("id #{} = {}".format(id, x))

Output:

#    id #0 = 8
#    id #1 = 23
#    id #2 = 45

We can access the indices of a pandas Series in a for loop using .items():

for id, x in xs.items():
    print("id #{} = {}".format(id, x))

Output:

#    id #G923002 = 8
#    id #G923004 = 23
#    id #G923005 = 45

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