Set value for particular cell in pandas DataFrame with iloc

Question

I have a question similar to this and this  The difference is that I have to select row by position  as I do not know the index    I want to do something like df iloc 0   COL NAME     x  but iloc does not allow this kind of access  If I do df iloc 0   COL NAME    x the warning about chained indexing appears

User · Answer

To modify the value in a cell at the intersection of row "r" (in column "A") and column "C"

retrieve the index of the row "r" in column "A"

    i = df[ df['A']=='r' ].index.values[0]

modify the value in the desired column "C"
```
    df.loc[i,"C"]="newValue"
```

Note: before, be sure to reset the index of rows ...to have a nice index list!

        df=df.reset_index(drop=True)

User · Answer

Another way is to get the row index and then use df loc or df at        get row index  label  from row number  irow  label   df index values irow   df at label   COL NAME     x

User · Answer

One thing I would add here is that the at function on a dataframe is much faster particularly if you are doing a lot of assignments of individual  not slice  values    df at index   col name     x   In my experience I have gotten a 20x speedup  Here is a write up that is Spanish but still gives an impression of what s going on

User · Answer

For mixed position and index  use  ix  BUT you need to make sure that your index is not of integer  otherwise it will cause confusions  df ix 0   COL NAME     x  Update  Alternatively  try df iloc 0  df columns get loc  COL NAME      x  Example  import pandas as pd import numpy as np    your data                            np random seed 0  df   pd DataFrame np random randn 10  2   columns   col1    col2    index np random randint 1 100 10   sort index    print df          col1    col2 10  1 7641  0 4002 24  0 1440  1 4543 29  0 3131 -0 8541 32  0 9501 -0 1514 33  1 8676 -0 9773 36  0 7610  0 1217 56  1 4941 -0 2052 58  0 9787  2 2409 75 -0 1032  0 4106 76  0 4439  0 3337     iloc with get loc                                       df iloc 0  df columns get loc  col2      100  df        col1      col2 10  1 7641  100 0000 24  0 1440    1 4543 29  0 3131   -0 8541 32  0 9501   -0 1514 33  1 8676   -0 9773 36  0 7610    0 1217 56  1 4941   -0 2052 58  0 9787    2 2409 75 -0 1032    0 4106 76  0 4439    0 3337

User · Answer

Extending Jianxun s answer  using set value mehtod in pandas  It sets value for a column at given index   From pandas documentations      DataFrame set value index  col  value    To set value at particular index for a column  do   df set value index   COL NAME   x    Hope it helps

User · Answer

another way is   you assign a column value for a given row based on the index position of a row  the index position always starts with zero  and the last index position is the length of the dataframe  df  quot COL NAME quot   iloc 0  x

User · Answer

You can use   df set value  Row index    Column name   value    set valye is  100 times faster than  ix method  It also better then use  df  Row index    Column name     value    But since set value is deprecated now so  iat  at are good replacements    For example if we have this data frame      A   B   C 0  1   8   4  1  3   9   6 2  22 33  52   if we want to modify the value of the cell  0  A   we can do   df iat 0 0    2   or df at 0  A     2

User · Answer

If you know the position  why not just get the index from that   Then use  loc   df loc index   COL NAME     x

[python] Set value for particular cell in pandas DataFrame with iloc

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