Python 基于现有列上的条件创建新列

我有一个数据框，如下所示：

col1 = ['a','b','c','a','c','a','b','c','a']
col2 = [1,1,0,1,1,0,1,1,0]
df2 = pd.DataFrame(zip(col1,col2),columns=['name','count'])

    name    count
0   a       1
1   b       1
2   c       0
3   a       1
4   c       1
5   a       0
6   b       1
7   c       1
8   a       0

我试图找出“name”列中每个元素对应的0个数与0+1之和的比率。 首先，我将计数加总如下：

for j in df2.name.unique():
    print(j)
    zero_ct = zero_one_frequencies[zero_one_frequencies['name'] == j][0]
    full_ct = zero_one_frequencies[zero_one_frequencies['name'] == j][0] + zero_one_frequencies[zero_one_frequencies['name'] == j][1]
    zero_pb = zero_ct / full_ct
    one_pb = 1 - zero_pb
    print(f"ZERO rations for {j} = {zero_pb}")
    print(f"One ratios for {j} = {one_pb}")
    print("="*30)

输出如下所示：

a
ZERO ratios for a = 0    0.5
dtype: float64
One ratios for a = 0    0.5
dtype: float64
==============================
b
ZERO ratios for b = 1    0.0
dtype: float64
One ratios for b = 1    1.0
dtype: float64
==============================
c
ZERO ratios for c = 2    0.333333
dtype: float64
One ratios for c = 2    0.666667
dtype: float64
==============================

我的目标是向数据框中添加两个新列：“name_0”和“name_1”，其中“name”列中的每个元素都有th比率值。我尝试了一些东西，但没有达到预期的效果：

for j in df2.name.unique():
    print(j)
    zero_ct = zero_one_frequencies[zero_one_frequencies['name'] == j][0]
    full_ct = zero_one_frequencies[zero_one_frequencies['name'] == j][0] + zero_one_frequencies[zero_one_frequencies['name'] == j][1]
    zero_pb = zero_ct / full_ct
    one_pb = 1 - zero_pb
    print(f"ZERO Probablitliy for {j} = {zero_pb}")
    print(f"One Probablitliy for {j} = {one_pb}")
    print("="*30)
    
    condition1 = [ df2['name'].eq(j) & df2['count'].eq(0)]
    condition2 = [ df2['name'].eq(j) & df2['count'].eq(1)]
    choice1 = zero_pb.tolist()
    choice2 = one_pb.tolist()

    print(f'choice1 = {choice1}, choice2 = {choice2}')
    df2["name"+str("_0")] = np.select(condition1, choice1, default=0)
    df2["name"+str("_1")] = np.select(condition2, choice2, default=0)

该列将使用name元素“c”的值进行更新。这是预期的，因为最后计算的值将用于更新所有值

你能帮我理解是否有其他有效使用np.select的方法吗

预期产出：

    name    count   name_0      name_1
0   a       1       0.000000    0.500000
1   b       1       0.000000    1.000000
2   c       0       0.333333    0.000000
3   a       1       0.000000    0.500000
4   c       1       0.000000    0.666667
5   a       0       0.500000    0.000000
6   b       1       0.000000    1.000000
7   c       1       0.000000    0.666667
8   a       0       0.500000    0.000000

---

下面的代码修复了这个问题。但是，我无法找到使用numpy.select获得相同结果的方法

df2["name"+str("_0")] = 0.0
df2["name"+str("_1")] = 0.0
for j in df2.name.unique():
    print(j)
    zero_ct = zero_one_frequencies[zero_one_frequencies['name'] == j][0]
    full_ct = zero_one_frequencies[zero_one_frequencies['name'] == j][0] + zero_one_frequencies[zero_one_frequencies['name'] == j][1]
    zero_pb = zero_ct / full_ct
    one_pb = 1 - zero_pb
    print(f"ZERO Probablitliy for {j} = {zero_pb.tolist()[0]}")
    print(f"One Probablitliy for {j} = {one_pb.tolist()[0]}")
    print("="*30)
    for idx in df2[df2['name']== j ].index:
        print("Index:::", idx)
        if df2['count'].iloc[idx] == 0:
            df2.at[idx, "name"+str("_0")] = zero_pb.tolist()[0]
            print(f'Count for {j} at index {idx} is {a}')
            print('printing name_0: ', df2["name"+str("_0")].iloc[idx])
            print("*"*30)
        elif df2['count'].iloc[idx] == 1:
            df2.at[idx, "name"+str("_1")] = one_pb.tolist()[0]
            print(f'Count for {j} at index {idx} is {b}')
            print('printing name_1: ', df2["name"+str("_1")].iloc[idx])
            print("*"*30)

---

我没有访问零度一度频率df的权限。所以我冒昧地尝试用我的方式解决这个问题

import pandas as pd
import numpy as np
col1 = ['a','b','c','a','c','a','b','c','a']
col2 = [1,1,0,1,1,0,1,1,0]
df2 = pd.DataFrame(zip(col1,col2),columns=['name','count'])

df2["name_0"] = 0
df2["name_1"] = 0

for name in df2['name'].unique():
  df_name = df2[df2['name'] == name]
  prob_1 = sum(df_name['count']/df_name.shape[0])
  for count in df2['count'].unique():
    indx = np.where((df2['name'] == name) & (df2['count'] == count))
    df2["name_" + str(count)].loc[indx] = np.abs(((count +1) % 2) - prob_1)

输出：

name    count   name_0  name_1
0   a   1   0.000000    0.500000
1   b   1   0.000000    1.000000
2   c   0   0.333333    0.000000
3   a   1   0.000000    0.500000
4   c   1   0.000000    0.666667
5   a   0   0.500000    0.000000
6   b   1   0.000000    1.000000
7   c   1   0.000000    0.666667
8   a   0   0.500000    0.000000

---

为了理解np。请选择我建议看。

请根据

df2

发布您的预期输出。嗨，梅亚克：编辑我的帖子以获得更好的澄清。关于清晰的代码@Oddaspa，请点击链接。它看起来比我的要干净得多：）`zero\u one\u frequencies=pd.crosstab（df2['name'，df2['count']）\.reset\u index（）.rename（columns={'index'：'count'}）\.rename\u axis（None，axis='columns'）`这是我使用的zero\u one\u frequencies代码