How to access column via index when using iterrows()

Question

I want to know how I can access columns using index rather than name when using iterrows to traverse DataFrames.

This code is most I could find:

for index, row in df.iterrows():
    print row['Date']

This is another approach I took to traverse, but it seems very slow:

for i in df.index:
    for j in range(len(df.columns)):       
                    df.ix[i,j] = 0

Answer 1

You can use ix to access by index:

In [67]: df
Out[67]:
       A  B
0  test1  1
1  test2  4
2  test3  1
3  test4  2

In [68]: df.ix[:,1]
Out[68]:
0    1
1    4
2    1
3    2
Name: B, dtype: int64

Updating your code with first column:

for index, row in df.iterrows():
    row.ix[0]

Answer 2

I figured it out. Iterate for i to number of columns and use i as index to access columns:

for i in range(len(df.columns)):  
    for index, row in df.iterrows():    
        print row.ix[i]