Wrt

The best thing that I can figure out is to do df.at[df.index[0],'oranges'], but that seems so awkward when they've gone out of their way to provide both by-label and by-integer-offset interfaces. Is that the best thing?

Yes, it is. And I agree, it is awkward. The old .ix used to support these mixed indexing cases better but its behaviour depended on the dtype of the axis, making it inconsistent. In the meanwhile...

The other options, which have been used in the other answers, can all issue the SettingWithCopy warning. It's not guaranteed to raise the issue but it might, based on what the indexing criteria are and how values are assigned.

Referencing Combining positional and label-based indexing and starting with this df, which has dateindex as the index:

                             apples oranges  bananas
dateindex
2021-01-01 14:00:01.384624        1      X         3
2021-01-05 13:43:26.203773        4      5         6
2021-01-31 08:23:29.837238        7      8         9
2021-02-08 10:23:09.095632        0      1         2

Using both options:

1. with .loc or .at:

   df.at[df.index[0], 'oranges'] = -50
   

                                apples oranges  bananas
   dateindex
   2021-01-01 14:00:01.384624        1     -50        3
   2021-01-05 13:43:26.203773        4      5         6
   2021-01-31 08:23:29.837238        7      8         9
   2021-02-08 10:23:09.095632        0      1         2
   

2. with .iloc or .iat:

   df.iat[0, df.columns.get_loc('oranges')] = -20
   

                                apples oranges  bananas
   dateindex
   2021-01-01 14:00:01.384624        1     -20        3
   2021-01-05 13:43:26.203773        4      5         6
   2021-01-31 08:23:29.837238        7      8         9
   2021-02-08 10:23:09.095632        0      1         2
   

FWIW, I find approach #1 more consistent since it can handle multiple row indexes without changing the functions/methods used: df.loc[df.index[[0, 2]], 'oranges'] but approach #2 needs a different column indexer when there are multiple columns: df.iloc[[0, 2], df.columns.get_indexer(['oranges', 'bananas'])].

Solution with Series.iat

If it doesn't seem more awkward to you, you can use the iat method of pandas Series:

df["oranges"].iat[0] = 2

Time performance comparison with other methods

As this method doesn't raise any warning, it can be interesting to compare its time performance with other proposed solutions.

%%timeit
df.at[df.index[0], 'oranges'] = 2
# > 9.91 µs ± 47.4 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
df.iat[0, df.columns.get_loc('oranges')] = 2
# > 13.5 µs ± 74.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
df["oranges"].iat[0] = 2
# > 3.49 µs ± 16.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

The pandas.Series.iat method seems to be the most performant one (I took the median of three runs).

Let's try again with huge DataFrames

With a DatetimeIndex

# Generating random data
df_large = pd.DataFrame(np.random.randint(0, 50, (100000, 100000)))
df_large.columns = ["col_{}".format(i) for i in range(100000)]
df_large.index = pd.date_range(start=0, periods=100000)
# 2070-01-01 to 2243-10-16, a bit unrealistic

%%timeit
df_large.at[df_large.index[55555], 'col_55555'] = -2
# > 10.1 µs ± 85.4 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
df_large.iat[55555, df_large.columns.get_loc('col_55555')] = -2
# > 13.2 µs ± 118 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
df_large["col_55555"].iat[55555] = -2
# > 3.31 µs ± 19 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

With a RangeIndex

# Generating random data
df_large = pd.DataFrame(np.random.randint(0, 50, (100000, 100000)))
df_large.columns = ["col_{}".format(i) for i in range(100000)]

%%timeit
df_large.at[df_large.index[55555], 'col_55555'] = 2
# > 4.5 µs ± 18.4 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
df_large.iat[55555, df_large.columns.get_loc('col_55555')] = 2
# > 13.5 µs ± 50.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
df_large["col_55555"].iat[55555] = 2
# > 3.49 µs ± 20.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Since it is a simple indexing with O(n) complexity, the size of the array doesn't change much the results, except when it comes to the "at + index" ; strangely enough, it shows worst performance with small dataframes. Thanks to the author wfaulk for spotting that using a RangeIndex decreases the access time of the "at + index" method. The time performance remains higher and constant when dealing with DatetimeIndex with pd.Series.iat.

You were actually quite close with your initial guess.

You would do it like this:

import pandas as pd

mydict = [{'a': 1, 'b': 2, 'c': 3, 'd': 4},
          {'a': 100, 'b': 200, 'c': 300, 'd': 400},
          {'a': 1000, 'b': 2000, 'c': 3000, 'd': 4000 }]
df = pd.DataFrame(mydict)
print(df)

# change th value of column a, row 2
df['a'][2] = 100

# print column a, row 2
print(df['a'][2])

There are lots of different variants such as loc and iloc, but this is one good method.

In the example we discovered that loc was optimal as df[][] throws an error:

import pandas as pd
data = [{'apples':1, 'oranges':'X', 'bananas':3},
        {'apples':4, 'oranges':5,   'bananas':6},
        {'apples':7, 'oranges':8,   'bananas':9},
        {'apples':0, 'oranges':1,   'bananas':2}]
indexes = [pd.to_datetime('2021-01-01 14:00:01.384624'),
           pd.to_datetime('2021-01-05 13:43:26.203773'),
           pd.to_datetime('2021-01-31 08:23:29.837238'),
           pd.to_datetime('2021-02-08 10:23:09.095632')]
idx = pd.Index(indexes, name='dateindex')
df = pd.DataFrame(data, index=idx)

print(df)

df.loc['2021-01-01 14:00:01.384624','oranges'] = 10
# df['oranges'][0] = 10

print(df)

This works.

You can use the loc method. It receives the row and column you want to change.

Changing X to 2: df.loc[0, 'oranges'] = 2

See: pandas.DataFrame.loc