Replace Na with 0 in a Data Frame Column

How to replace NaN values by Zeroes in a column of a Pandas Dataframe?

I believe DataFrame.fillna() will do this for you.

Link to Docs for a dataframe and for a Series.

Example:

In [7]: df
Out[7]: 
          0         1
0       NaN       NaN
1 -0.494375  0.570994
2       NaN       NaN
3  1.876360 -0.229738
4       NaN       NaN

In [8]: df.fillna(0)
Out[8]: 
          0         1
0  0.000000  0.000000
1 -0.494375  0.570994
2  0.000000  0.000000
3  1.876360 -0.229738
4  0.000000  0.000000

To fill the NaNs in only one column, select just that column. in this case I'm using inplace=True to actually change the contents of df.

In [12]: df[1].fillna(0, inplace=True)
Out[12]: 
0    0.000000
1    0.570994
2    0.000000
3   -0.229738
4    0.000000
Name: 1

In [13]: df
Out[13]: 
          0         1
0       NaN  0.000000
1 -0.494375  0.570994
2       NaN  0.000000
3  1.876360 -0.229738
4       NaN  0.000000

EDIT:

To avoid a SettingWithCopyWarning, use the built in column-specific functionality:

df.fillna({1:0}, inplace=True)

Replace NA with 0 in a data frame column

Since nobody so far felt fit to point out why what you're trying doesn't work:

1. NA == NA doesn't return TRUE, it returns NA (since comparing to undefined values should yield an undefined result).
2. You're trying to call apply on an atomic vector. You can't use apply to loop over the elements in a column.
3. Your subscripts are off - you're trying to give two indices into a$x, which is just the column (an atomic vector).

I'd fix up 3. to get to a$x[is.na(a$x)] <- 0

Python Pandas replace multiple columns zero to Nan

I think you need replace by dict:

cols = ["Weight","Height","BootSize","SuitSize","Type"]
df2[cols] = df2[cols].replace({'0':np.nan, 0:np.nan})

How do I replace NA values with zeros in an R dataframe?

See my comment in @gsk3 answer. A simple example:

> m <- matrix(sample(c(NA, 1:10), 100, replace = TRUE), 10)
> d <- as.data.frame(m)
   V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1   4  3 NA  3  7  6  6 10  6   5
2   9  8  9  5 10 NA  2  1  7   2
3   1  1  6  3  6 NA  1  4  1   6
4  NA  4 NA  7 10  2 NA  4  1   8
5   1  2  4 NA  2  6  2  6  7   4
6  NA  3 NA NA 10  2  1 10  8   4
7   4  4  9 10  9  8  9  4 10  NA
8   5  8  3  2  1  4  5  9  4   7
9   3  9 10  1  9  9 10  5  3   3
10  4  2  2  5 NA  9  7  2  5   5

> d[is.na(d)] <- 0

> d
   V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1   4  3  0  3  7  6  6 10  6   5
2   9  8  9  5 10  0  2  1  7   2
3   1  1  6  3  6  0  1  4  1   6
4   0  4  0  7 10  2  0  4  1   8
5   1  2  4  0  2  6  2  6  7   4
6   0  3  0  0 10  2  1 10  8   4
7   4  4  9 10  9  8  9  4 10   0
8   5  8  3  2  1  4  5  9  4   7
9   3  9 10  1  9  9 10  5  3   3
10  4  2  2  5  0  9  7  2  5   5

There's no need to apply apply. =)

EDIT

You should also take a look at norm package. It has a lot of nice features for missing data analysis. =)

Replace NA in DataFrame for multiple columns with mean per country

Note Example is based on your additional data source from the comments

Replacing the NA-Values for multiple columns with mean() you can combine the following three methods:

• fillna() (Iterating per column axis should be 0, which is default value of fillna())
• groupby()
• transform()

Create data frame from your example:

df = pd.read_excel('https://happiness-report.s3.amazonaws.com/2021/DataPanelWHR2021C2.xls')

library(dplyr)
library(tidyr)
df1 <- df1 %>%
         mutate(across(contains('keyword'), replace_na, 0))

df1
# A tibble: 5 × 4
   col1 col2_keyword col3   col4
  <int> <chr>        <chr> <dbl>
1     1 a            a         1
2     2 b            b         3
3     3 0            c        NA
4     4 c            d         5
5     5 d            <NA>      6

df <- df %>%
    mutate(across(where(~ is.character(.x) && 'keyword' %in% .x), replace_na, 0))

df
# A tibble: 5 × 4
   col1 col2    col3   col4
  <int> <chr>   <chr> <dbl>
1     1 a       a         1
2     2 b       b         3
3     3 keyword c        NA
4     4 0       d         5
5     5 c       <NA>      6

df <- tibble(col1 = 1:5, col2 = c("a", "b", "keyword", NA, 'c'), 
 col3 = c('a', 'b', 'c', 'd', NA), col4 = c(1, 3, NA, 5, 6))
df1 <- tibble(col1 = 1:5, col2_keyword = c("a", "b", NA, 'c', 'd'),
    col3 =c('a', 'b', 'c', 'd', NA), col4 = c(1, 3, NA, 5, 6))

library(tidyverse)
df1 <- tibble(x = seq(3), y = c(1, NA, 2))
df1 %>% mutate(y = y %>% replace_na(mean(df1$y, na.rm = TRUE)))
#> # A tibble: 3 × 2
#>       x     y
#>   <int> <dbl>
#> 1     1   1  
#> 2     2   1.5
#> 3     3   2