R help - Aug 2021 - substitute column data frame based on name stored in variable in r

Hello,
I would like to recursively select the columns of a dataframe by
strong the names of the dataframe in a vector and extracting one
element of the vector at a time. This I can do with, for instance:
```
vect = names(df)
sub_df[vect[1]]
```

The problem is that I would like also to change the values of the
selected column using some logic as in `df$column[df$column == value]
<- new.value`, but I am confused on the syntax for the vectorized
version. Specifically, this does not work:
```
sub_df[vect[1] == 0] = "No"
```
What would be the correct approach?
Thank you

Hi Luigi,
It looks to me as though you will have to copy the data frame or store
the output in a new data frame.

Jim

On Mon, Aug 9, 2021 at 6:26 PM Luigi Marongiu <marongiu.luigi at
gmail.com> wrote:
>
> Hello,
> I would like to recursively select the columns of a dataframe by
> strong the names of the dataframe in a vector and extracting one
> element of the vector at a time. This I can do with, for instance:
> ```
> vect = names(df)
> sub_df[vect[1]]
> ```
>
> The problem is that I would like also to change the values of the
> selected column using some logic as in `df$column[df$column == value]
> <- new.value`, but I am confused on the syntax for the vectorized
> version. Specifically, this does not work:
> ```
> sub_df[vect[1] == 0] = "No"
> ```
> What would be the correct approach?
> Thank you
>
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On Mon, 9 Aug 2021 10:26:03 +0200
Luigi Marongiu <marongiu.luigi at gmail.com> wrote:
> vect = names(df)
> sub_df[vect[1]]
> df$column[df$column == value] <- new.value
Let's see, an equivalent expression without the $ syntax is
`df[['column']][df[['column']] == value] <- new.value`.
Slightly
shorter, matrix-like syntax would give us
`df[df[['column']] == value, 'column'] <- new.value`.

Now replace 'column' with vect[i] and you're done. The `[[`-indexing
is
used here to get the column contents instead of a single-column
data.frame that `[`-indexing returns for lists.

Also note that df[[names(df)[i]]] should be the same as df[[i]] for
most data.frames.

-- 
Best regards,
Ivan