如何将公式应用于 Dataframe pandas 中的所有列

Question

I have the following Dataframe:

import pandas as pd
data = {'MA1': [ float("nan"),  float("nan"),      -1,   1],
        'MA2': [ float("nan"),            -1,       0,   0],
        'MA3': [            0,             0,       1,  -1]}
df_input = pd.DataFrame(data, columns=['MA1', 'MA2', 'MA3'])

My goal is for every column, if the first non nan and non zero value is -1, to set it to 0.

Clarification:

The goal is only to set to 0 if the first non 0 and non nan value is -1. If it is 1 or anything else, then leave it there.

What is the fastest way to do it?

Answer 1

You can loop over the columns and use DataFrame.loc to assign the 0 when the first valid value is -1 :

dft = df_input.replace(0, np.NaN)

for col in df_input.columns:
    idxmin = dft[col].idxmin()
    if df_input.loc[idxmin, col] == -1:
        df_input.loc[idxmin, col] = 0

   MA1  MA2  MA3
0  NaN  NaN    0
1  NaN  0.0    0
2  0.0  0.0    1
3  1.0  0.0    0

---

Or more efficient by using DataFrame.idxmin instead so we dont have to to call Series.idxmin for each iteration in our loop:

dft = df_input.replace(0, np.NaN).idxmin()

for col, idx in dft.iteritems():
    if df_input.loc[idx, col] == -1:
        df_input.loc[idx, col] = 0

   MA1  MA2  MA3
0  NaN  NaN    0
1  NaN  0.0    0
2  0.0  0.0    1
3  1.0  0.0    0

Answer 2

Being at the end of one year using python, I'm trying to be better at implementing higher performing solutions, so I thought I would test the performance of my answer versus other's (realizing that my answer would be the slowest -- from the dataframe I created , it ended up being 50,000x slower than the best answer. Woah,): Also, here is a good article about pandas and performance: https://engineering.upside.com/a-beginners-guide-to-optimizing-pandas-code-for-speed-c09ef2c6a4d6

My traditional slow looping method looped through 3 columns almost 100,000 times (length of dataframe), while the best answer looped through 3 columns one time as it idx.min() identified the relevant row, making it unnecessary to loop through them all.

Here is a dataframe with 100,000 rows and 4 columns that I used to test vs. @Erfan and @DerekO:

df_input = pd.DataFrame(np.random.randint(0, 10, size=(100000,4)).astype(float), columns=list('ABCD'))
df_input.iloc[99998:, 0:4] = -1

My Answer (slowest) 2.78 s ± 269 ms per loop :

for col in df_input.columns:
    for row in range(len(df_input.index)):
        if df_input.loc[row, col] == -1:
            df_input.loc[row, col] = 0
            break    
df_input

Derek O's answer #1: 283 ms ± 13.2 ms per loop 10x faster than my answer!

Erfan's answer #1: 2.73 ms ± 135 µs per loop 1,000x faster than my answer!

Erfan's answer #2: 54.8 µs ± 5.65 µs per loop 50,000x faster than my answer!

Answer 3

Apply a custom function to each column. The custom function loops through the column's values to find the first non-nan, non-zero value, then returns the new column.

import numpy as np
import pandas as pd

def set_column(col_values):
    for index, value in enumerate(col_values):
        if value != 0 and not np.isnan(value):
            if value == -1:
                col_values[index] = 0
                return col_values
            else:
                return col_values

data = {'MA1': [ float("nan"),  float("nan"),      -1,   1],
        'MA2': [ float("nan"),            -1,       0,   0],
        'MA3': [            0,             0,       1,   0]}

df_input = pd.DataFrame(data, columns=['MA1', 'MA2', 'MA3'])
df_output = df_input.copy().apply(lambda x: set_column(x), axis = 0)

Output:

>>> df_output
   MA1  MA2  MA3
0  NaN  NaN    0
1  NaN  0.0    0
2  0.0  0.0    1
3  1.0  0.0    0

Answer 4

I used a modification of @Erfan's answer.

As I explain in my Update edit, I want to only set it to zero if the first non zero and non nan value is -1. If it anything else, then don't do anything for that column.

df_min = df_input(0, np.NaN).idxmin()
df_max = df_input(0, np.NaN).idxmax()
for col, idx in df_min.iteritems():
    if df_input[idx, col] == -1 and idx < df_max[col]:
        df_input[idx, col] = 0