计算 Pandas dataframe 的每一列中满足条件的值的数量

Question

I have a dataframe with several columns of data. In the data, a -1 is equivalent to missing data. I want to count the number of -1 values in each column.

I believe I could register -1 as a NaN/missing value when I load the data and then I saw something that used isna() and counted boolean values. However, what I want to do (apply a condition to each column) seems like a fundamental thing I should know how to do, so I would like to figure out how to do it this way.

Here is an example. Imagine I have the following data frame:

row   A   B  C  D  E
1     3   5  6  9 -1
2    -1   3 -1  2  0
3    -1  -1 -1  1 -1

The output I would like to get would be:

A  B  C  D  E
2  1  2  0  2

I have tried the following:

df.apply(lambda x: x == -1).count() # value returned was the count of all the rows
(df == -1).count() # also returned a count of all the rows.

I looked through several questions related to "countif", but they all seemed to apply a condition to one column to select rows. And the two items I tried above were from questions related to apply functions to each column and count values that match a condition in each column.

The suggested duplicate in the comments is looking for a single value for the entire dataframe and different criteria on each column. I am looking to apply the same condition to every column and get a result per column, as shown in the selected answer below.

I would appreciate any thoughts or ideas on how to proceed.

Answer 1

Use DataFrame.eq + DataFrame.sum :

#You can omit to_frame and T if you don't want a DataFrame.
df.eq(-1).sum().to_frame().T
#(df==-1).sum() #similar

or if it is str :

df.eq('-1').sum().to_frame().T

---

if row is a column:

df[df.columns[1:]].eq(-1).sum().to_frame().T

   A  B  C  D  E
0  2  1  2  0  2