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Pandas Dataframe:创建一个新列,并根据其他列上的2个条件语句填充值

conditional-statements pandas

Pablo · 技术社区 · 1 年前

我编写了这个脚本,它根据满足两个条件的值创建新列。

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df=pd.DataFrame()

df['variable 1']= np.arange(0,1.1,0.1)
df['variable 2']= 0.2*df['variable 1']
df['variable 3']= 0.4 -0.2*df['variable 1']


# Create new columns 

slope = [2, 1.5, 1, 0.5]

for i in range(len(slope)):

    df['slope = ' + str(slope[i])]=''
    for j in range(len(df['variable 1'])):
    # Calculating Scl_disp_sd with equation 1
        curve = 0.5 - slope[i]*df['variable 1'][j]
        df['slope = ' + str(slope[i])][j]= np.where((curve>df['variable 2'][j]) & (curve<df['variable 3'][j]), curve,np.nan)

display(df)

plt.plot(df['variable 1'], df['variable 2'], 'o', label='variable 2')
plt.plot(df['variable 1'], df['variable 3'], 'o', label='variable 3')
plt.plot(df['variable 1'], df.filter(like='slope =', axis=1), marker='.')
plt.legend()

enter image description here

这个脚本是有效的,但我收到了以下消息:

/var/folders/m0/_y1fs5x50xx99pjg2yf42y7r0000gp/T/ipykernel_1964/2618301266.py:11: FutureWarning: ChainedAssignmentError: behaviour will change in pandas 3.0!
You are setting values through chained assignment. Currently this works in certain cases, but when using Copy-on-Write (which will become the default behaviour in pandas 3.0) this will never work to update the original DataFrame or Series, because the intermediate object on which we are setting values will behave as a copy.
A typical example is when you are setting values in a column of a DataFrame, like:

df["col"][row_indexer] = value

Use `df.loc[row_indexer, "col"] = values` instead, to perform the assignment in a single step and ensure this keeps updating the original `df`.

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy

  df['slope = ' + str(slope[i])][j]= np.where((curve>df['variable 2'][j]) & (curve<df['variable 3'][j]),
/var/folders/m0/_y1fs5x50xx99pjg2yf42y7r0000gp/T/ipykernel_1964/2618301266.py:11: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
  df['slope = ' + str(slope[i])][j]= np.where((curve>df['variable 2'][j]) & (curve<df['variable 3'][j]),
/var/folders/m0/_y1fs5x50xx99pjg2yf42y7r0000gp/T/ipykernel_1964/2618301266.py:11: FutureWarning: ChainedAssignmentError: behaviour will change in pandas 3.0!
You are setting values through chained assignment. Currently this works in certain cases, but when using Copy-on-Write (which will become the default behaviour in pandas 3.0) this will never work to update the original DataFrame or Series, because the intermediate object on which we are setting values will behave as a copy.
A typical example is when you are setting values in a column of a DataFrame, like:

df["col"][row_indexer] = value

Use `df.loc[row_indexer, "col"] = values` instead, to perform the assignment in a single step and ensure this keeps updating the original `df`.

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
...
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
  df['slope = ' + str(slope[i])][j]= np.where((curve>df['variable 2'][j]) & (curve<df['variable 3'][j]),
Output is truncated. View as a scrollable element or open in a text editor. Adjust cell output settings...

如果有人能想出另一个办法来写这个脚本,以避免这个信息,我将不胜感激

1 回复 | 直到 1 年前

mozway 1 年前

不需要嵌套循环。只需将操作作为向量应用:

slope = [2, 1.5, 1, 0.5]

for i in range(len(slope)):
    curve = 0.5 - slope[i]*df['variable 1']
    df['slope = ' + str(slope[i])] = np.where((curve>df['variable 2'])
                                               & (curve<df['variable 3']),
                                              curve,np.nan)

或完全矢量化 numpy :

curve = 0.5 - slope*df['variable 1'].to_numpy()[:, None]
cols = [f'slope = {c}' for c in slope]
df[cols] = np.where(  (curve > df[['variable 2']].to_numpy())
                    & (curve < df[['variable 3']].to_numpy()),
                    curve, np.nan)

输出:

    variable 1  variable 2  variable 3  slope = 2  slope = 1.5  slope = 1  slope = 0.5
0          0.0        0.00        0.40        NaN          NaN        NaN          NaN
1          0.1        0.02        0.38        0.3         0.35        NaN          NaN
2          0.2        0.04        0.36        0.1         0.20        0.3          NaN
3          0.3        0.06        0.34        NaN          NaN        0.2          NaN
4          0.4        0.08        0.32        NaN          NaN        0.1         0.30
5          0.5        0.10        0.30        NaN          NaN        NaN         0.25
6          0.6        0.12        0.28        NaN          NaN        NaN         0.20
7          0.7        0.14        0.26        NaN          NaN        NaN         0.15
8          0.8        0.16        0.24        NaN          NaN        NaN          NaN
9          0.9        0.18        0.22        NaN          NaN        NaN          NaN
10         1.0        0.20        0.20        NaN          NaN        NaN          NaN