在循环 dataframe 时抓取 pandas 列的更新行

Question

I am trying the following:

import pandas as pd
df = pd.DataFrame({'Col1': {0: 'A', 1: 'A', 2: 'B', 3: 'B', 4: 'B'},
 'Col2': {0: 'a', 1: 'a', 2: 'b', 3: 'b', 4: 'c'},
 'Col3': {0: 42, 1: 28, 2: 56, 3: 62, 4: 48}})

ii = 1
for idx, row in df.iterrows():
    print(row)
    df.at[:, 'Col2'] = 'asd{}'.format(ii)
    ii += 1

But the print statement above doesn't reflect the change df.at[:, 'Col2'] = 'asd'.format(ii) . I need the print statements to reflect the change df.at[:, 'Col2'] = 'asd'.format(ii)

Edit: Since I am updating all rows of df , I was expecting the idx and row to grab new values from dataframe.

If this is not the right way to grab updated values from df through idx and row , then what is the correct approach. I need idx and row to reflect new values.

Expected output:

Col1     A
Col2     a
Col3    42
Name: 0, dtype: object
Col1     A
Col2     asd1
Col3    28
Name: 1, dtype: object
Col1     B
Col2     asd2
Col3    56

.....

Answer 1

From iterrows documentation :

You should never modify something you are iterating over. This is not guaranteed to work in all cases. Depending on the data types, the iterator returns a copy and not a view, and writing to it will have no effect.

As per your request for an alternative solution, here is one using DataFrame.apply :

df['Col2'] = df.apply(lambda row: 'asd{}'.format(row.name), axis=1)

Other examples (also using Series.apply ) that may be useful for your eventual goal: (not clear what it is yet)

df['Col2'] = df['Col2'].apply(lambda x: 'asd{}'.format(x))
df['Col2'] = df.apply(lambda row: 'asd{}'.format(row['Col3']), axis=1)

Answer 2

Here is something you can try,

import pandas as pd

df = pd.DataFrame({'Col1': {0: 'A', 1: 'A', 2: 'B', 3: 'B', 4: 'B'},
                   'Col2': {0: 'a', 1: 'a', 2: 'b', 3: 'b', 4: 'c'},
                   'Col3': {0: 42, 1: 28, 2: 56, 3: 62, 4: 48}})

print(
    df.assign(idx=df.index)[['idx', 'Col2']]
        .apply(lambda x: x['Col2'] if x['idx'] == 0 else f"asd{x['idx']}", axis=1)
)

---

0       a
1    asd1
2    asd2
3    asd3
4    asd4
dtype: object