我有一个数据框:
Type: Volume:
Q 10
Q 20
T 10
Q 10
T 20
T 20
Q 10
我希望将T型组合成一行,只有当两个(或更多)Ts连续时才加总音量
即。致:
Q 10
Q 20
T 10
Q 10
T 20+20=40
Q 10
有没有办法实现这个目标?会DataFrame.groupby
工作吗?
答案 0 :(得分:1)
如果你只需要部分总和,这里有一个小技巧:
import numpy as np
import pandas as pd
df = pd.DataFrame({"Type": ["Q", "Q", "T", "Q", "T", "T", "Q"],
"Volume": [10, 20, 10, 10, 20, 20, 10]})
s = np.diff(np.r_[0, df.Type == "T"])
s[s < 0] = 0
res = df.groupby(("Type", np.cumsum(s) - 1)).sum().loc["T"]
print(res)
输出:
Volume
0 10
1 40
答案 1 :(得分:1)
我认为这会有所帮助。此代码可以处理任意数量的连续'T',您甚至可以更改要组合的字符。我在代码中添加了注释来解释它的作用。
import pandas as pd
def combine(df):
combined = [] # Init empty list
length = len(df.iloc[:,0]) # Get the number of rows in DataFrame
i = 0
while i < length:
num_elements = num_elements_equal(df, i, 0, 'T') # Get the number of consecutive 'T's
if num_elements <= 1: # If there are 1 or less T's, append only that element to combined, with the same type
combined.append([df.iloc[i,0],df.iloc[i,1]])
else: # Otherwise, append the sum of all the elements to combined, with 'T' type
combined.append(['T', sum_elements(df, i, i+num_elements, 1)])
i += max(num_elements, 1) # Increment i by the number of elements combined, with a min increment of 1
return pd.DataFrame(combined, columns=df.columns) # Return as DataFrame
def num_elements_equal(df, start, column, value): # Counts the number of consecutive elements
i = start
num = 0
while i < len(df.iloc[:,column]):
if df.iloc[i,column] == value:
num += 1
i += 1
else:
return num
return num
def sum_elements(df, start, end, column): # Sums the elements from start to end
return sum(df.iloc[start:end, column])
frame = pd.DataFrame({"Type": ["Q", "Q", "T", "Q", "T", "T", "Q"],
"Volume": [10, 20, 10, 10, 20, 20, 10]})
print(combine(frame))