pandas 如何在熊猫中填充重复数据的行？
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How to fill in rows with repeating data in pandas?
提问by Amyunimus
In R, when adding new data of unequal length to a data frame, the values repeat to fill the data frame:
在 R 中，当向数据帧添加不等长的新数据时，这些值会重复以填充数据帧：
df <- data.frame(first=c(1,2,3,4,5,6))
df$second <- c(1,2,3)
yielding:
产生：
first second
1 1 1
2 2 2
3 3 3
4 4 1
5 5 2
6 6 3
However, pandas requires equal index lengths.
但是，pandas 需要相同的索引长度。
How do I "fill in" repeating data in pandas like I can in R?
如何像在 R 中一样在 Pandas 中“填写”重复数据？
采纳答案by Yeqing Zhang
Seems there is no elegant way. This is the workaround I just figured out. Basically create a repeating list just bigger than original dataframe, and then left join them.
似乎没有优雅的方式。这是我刚刚想出的解决方法。基本上创建一个比原始数据框大的重复列表，然后将它们加入。
import pandas
df = pandas.DataFrame(range(100), columns=['first'])
repeat_arr = [1, 2, 3]
df = df.join(pandas.DataFrame(repeat_arr * (len(df)/len(repeat_arr)+1),
columns=['second']))
回答by Meow
The cycle method from itertools is good for repeating a common pattern.
itertools 中的循环方法非常适合重复常见的模式。
from itertools import cycle
seq = cycle([1, 2, 3])
df['Seq'] = [next(seq) for count in range(df.shape[0])]
回答by unutbu
import pandas as pd
import numpy as np
def put(df, column, values):
df[column] = 0
np.put(df[column], np.arange(len(df)), values)
df = pd.DataFrame({'first':range(1, 8)})
put(df, 'second', [1,2,3])
yields
产量
first second
0 1 1
1 2 2
2 3 3
3 4 1
4 5 2
5 6 3
6 7 1
Not particularly beautiful, but one "feature" it possesses is that you do not have to worry if the length of the DataFrame is a multiple of the length of the repeated values. np.putrepeats the values as necessary.
不是特别漂亮，但它拥有的一个“功能”是您不必担心 DataFrame 的长度是否是重复值长度的倍数。np.put根据需要重复这些值。
My first answer was:
我的第一个回答是：
import itertools as IT
df['second'] = list(IT.islice(IT.cycle([1,2,3]), len(df)))
but it turns out this is significantly slower:
但事实证明这要慢得多：
In [312]: df = pd.DataFrame({'first':range(10**6)})
In [313]: %timeit df['second'] = list(IT.islice(IT.cycle([1,2,3]), len(df)))
10 loops, best of 3: 143 ms per loop
In [316]: %timeit df['second'] = 0; np.put(df['second'], np.arange(N), [1,2,3])
10 loops, best of 3: 27.9 ms per loop
回答by Paul H
How general of a solution are you looking for? I tried to make this a little less hard-coded:
您正在寻找的解决方案有多普遍？我试图让这个不那么硬编码：
import numpy as np
import pandas
df = pandas.DataFrame(np.arange(1,7), columns=['first'])
base = [1, 2, 3]
df['second'] = base * (df.shape[0]/len(base))
print(df.to_string())
first second
0 1 1
1 2 2
2 3 3
3 4 1
4 5 2
5 6 3
回答by Daniele
In my case I needed to repeat the values without knowing the length of the sub-list, i.e. checking the length of every group. This was my solution:
在我的情况下，我需要在不知道子列表长度的情况下重复这些值，即检查每个组的长度。这是我的解决方案：
import numpy as np
import pandas
df = pandas.DataFrame(['a','a','a','b','b','b','b'], columns=['first'])
list = df.groupby('first').apply(lambda x: range(len(x))).tolist()
loop = [val for sublist in list for val in sublist]
df['second']=loop
df
first second
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 b 2
6 b 3
回答by SBM
Probably inefficient, but here's sort of a pure pandas solution.
可能效率低下，但这是一种纯Pandas解决方案。
import numpy as np
import pandas as pd
base = [1,2,3]
df = pd.DataFrame(data = None,index = np.arange(10),columns = ["filler"])
df["filler"][:len(base)] = base
df["tmp"] = np.arange(len(df)) % len(base)
df["filler"] = df.sort_values("tmp")["filler"].ffill() #.sort_index()
print(df)
回答by JDenman6
You might want to try using the power of modulo (%). You can take the value (or index) of first and use the length of second as the modulus to get the value (or index) you're looking for. Something like:
您可能想尝试使用模数 (%) 的幂。您可以取 first 的值（或索引）并使用 second 的长度作为模数来获取您要查找的值（或索引）。就像是：
df = pandas.DataFrame([0,1,2,3,4,5], columns=['first'])
sec = [0,1,2]
df['second'] = df['first'].apply(lambda x: x % len(sec) )
print(df)
first second
0 0 0
1 1 1
2 2 2
3 3 0
4 4 1
5 5 2
I hope that helps.
我希望这有帮助。
