熊猫数据框中的NaN值不变

Question

I have got a pandas dataframe which looks like this.

import pandas as pd
import math

inp = [{'c1':-5, 'c2':-2, 'c3':-3,'c4:360}, {'c1'1:, 'c2':-3, 'c3':4,'c4:550}, {'c1':NaN, 'c2':NaN, 'c3':NaN,'c4:NaN}]
df = pd.DataFrame(inp)

Output:

   c1  c2  c3  c4
0  -5  -2  -3  360
1   1  -3  4   550
2  NaN NaN NaN NaN

I want to itterate through the rows and change those who got an NaN value in c4. and then change the value in all of the cells which have got this.

for index, row in df.iterrows():
    if math.isnan(row["c4"]) == True:
    row["c1"] = float(0)
    row["c2"] = float(df["c1"][0]) + float(df["c2"][0])
    row["c3"] = row["c2"] - row["c1"]
    row["c4"] = float(2880)

For the first column it should just be 0.

For the second column it should be the first rows values in c1 and c2 added together.

For the third column should be the first columns minus the second column.

And the last column should just be 2880.

When i run the code above i do not get an error but the dataframe doesnt change at all. Can any1 tell me why or help me writing some code that does this?

Appreciate it!

Answer 1

You can avoid looping by using regular indexing:

idx = df.c4.isnull()
df.loc[idx,'c1'] = 0
df.loc[idx,'c2'] = df.iloc[0,0] + df.iloc[0,1]
df.loc[idx,'c3'] = df.loc[idx,'c2'] - df.loc[idx,'c1']
df.loc[idx,'c4'] = 2880.

Output:

 c1   c2   c3      c4
0 -5.0 -2.0 -3.0   360.0
1  1.0  3.0  4.0   550.0
2  0.0 -7.0 -7.0  2880.0