我有一個數據框，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 count0 a 11 b 12 c 03 a 14 c 15 a 06 b 17 c 18 a 0我試圖找到“名稱”列中每個元素對應的零數與零+一總和的比率。首先我將計數匯總如下：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)輸出如下：aZERO ratios for a = 0 0.5dtype: float64One ratios for a = 0 0.5dtype: float64==============================bZERO ratios for b = 1 0.0dtype: float64One ratios for b = 1 1.0dtype: float64==============================cZERO ratios for c = 2 0.333333dtype: float64One ratios for c = 2 0.666667dtype: float64==============================我的目標是向數據框中添加 2 個新列：“name_0”和“name_1”，以及“name”列中每個元素的比率值。我嘗試了一些方法，但沒有給出預期的結果：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()該列將使用名稱元素“c”的值進行更新。這是可以預料的，因為最后計算的值將用于更新所有值。還有另一種方法可以有效地使用 np.select 嗎？
2 回答
慕俠2389804
TA貢獻1719條經驗 獲得超6個贊
我無法訪問 Zero_one_frequencies df。所以我冒昧地嘗試用我的方式解決這個問題。
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)
輸出：
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
慕容3067478
TA貢獻1773條經驗 獲得超3個贊
以下代碼解決了該問題。但是，我找不到使用 numpy.select 獲得相同效果的方法。
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 ')
print('printing name_1: ', df2["name"+str("_1")].iloc[idx])
print("*"*30)
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