[英]Compare three columns and choose the highest
我有一個如下圖所示的數據集,
我的目標是比較最后三行並每次選擇最高的。
我有四個新變量:empty = 0、cancel = 0、release = 0、undertermined = 0
對於索引 0,cancelCount 是最高的,因此 cancel += 1。只有當三行相同時,未確定的才會增加。
這是我失敗的代碼示例:
empty = 0
cancel = 0
release = 0
undetermined = 0
if (df["emptyCount"] > df["cancelcount"]) & (df["emptyCount"] > df["releaseCount"]):
empty += 1
elif (df["cancelcount"] > df["emptyCount"]) & (df["cancelcount"] > df["releaseCount"]):
cancel += 1
elif (df["releasecount"] > df["emptyCount"]) & (df["releasecount"] > df["emptyCount"]):
release += 1
else:
undetermined += 1
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
一般來說,您應該避免循環。 這是一個滿足您需要的矢量化代碼示例:
# data of intereset
s = df[['emptyCount', 'cancelCount', 'releaseCount']]
# maximum by rows
max_vals = s.max(1)
# those are equal to max values:
equal_max = df.eq(max_vals, axis='rows').astype(int)
# If there are single maximum along the rows:
single_max = equal_max.sum(1)==1
# The values:
equal_max.mul(single_max, axis='rows').sum()
Output 將是一個如下所示的系列:
emmptyCount count1
cancelCount count2
releaseCount count3
dtype: int64
首先我們找到未確定的行
equal = (df['emptyCount'] == df['cancelcount']) | (df['cancelount'] == df['releaseCount'])
然后我們找到確定行的最大列
max_arg = df.loc[~equal, ['emptyCount', 'cancelcount', 'releaseCount']].idxmax(axis=1)
數一數
undetermined = equal.sum()
empty = (max_arg == 'emptyCount').sum()
cancel = (max_arg == 'cancelcount').sum()
release = (max_arg == 'releaseCount').sum()
import pandas as pd
import numpy as np
class thing(object):
def __init__(self):
self.value = 0
empty , cancel , release , undetermined = [thing() for i in range(4)]
dictt = { 0 : empty, 1 : cancel , 2 : release , 3 : undetermined }
df = pd.DataFrame({
'emptyCount': [2,4,5,7,3],
'cancelCount': [3,7,8,11,2],
'releaseCount': [2,0,0,5,3],
})
for i in range(1,4):
series = df.iloc[-4+i]
for j in range(len(series)):
if series[j] == series.max():
dictt[j].value +=1
cancel.value
獲取最大值的小腳本:
import numpy as np
emptyCount = [2,4,5,7,3]
cancelCount = [3,7,8,11,2]
releaseCount = [2,0,0,5,3]
# Here we use np.where to count instances where there is more than one index with the max value.
# np.where returns a tuple, so we flatten it using "for n in m"
count = [n for z in zip(emptyCount, cancelCount, releaseCount) for m in np.where(np.array(z) == max(z)) for n in m]
empty = count.count(0) # 1
cancel = count.count(1) # 4
release = count.count(2) # 1
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.