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[英]how to fill NaN values of Pandas column with values from another column
[英]Pandas Fill NaN with Column Values
给定以下数据框:
import pandas as pd
import numpy as np
df = pd.DataFrame({'A':[1,1,np.nan],
'B':[2.2,np.nan,2.2]})
df
A B
0 1.0 2.2
1 1.0 NaN
2 NaN 2.2
如果我想用在该列(1)中重复的值替换A列中的NaN值,并对B列执行相同的操作,我需要使用哪种fillna()?
A B
0 1.0 2.2
1 1.0 NaN
2 NaN 2.2
寻找通用解决方案,因为我确实有数千行。 提前致谢!
fillna
可以采用值的字典,其中键是列名。
假设您要用重复次数最多的值填充列,则可以使用以下方法计算字典:
df = pd.DataFrame({
'A': [1, 1, np.nan, 2],
'B': [2.2, np.nan, 2.2, 1.9]
})
fill_dict = df.mode().to_dict(orient='records')[0]
df = df.fillna(values=fill_dict)
df
A B
0 1 2.2
1 1 2.2
2 1 2.2
3 2 1.9
为什么不简单:
df.fillna(method='ffill')
# df = pd.DataFrame({'A': [1, 1, np.nan, 2], 'B': [2.2, np.nan, 2.2, 1.9]})
# df.fillna(method='ffill')
# A B
#0 1 2.2
#1 1 2.2
#2 1 2.2
#3 2 1.9
import itertools
import operator
def most_common(L):
# get an iterable of (item, iterable) pairs
SL = sorted((x, i) for i, x in enumerate(L))
# print 'SL:', SL
groups = itertools.groupby(SL, key=operator.itemgetter(0))
# auxiliary function to get "quality" for an item
def _auxfun(g):
item, iterable = g
count = 0
min_index = len(L)
for _, where in iterable:
count += 1
min_index = min(min_index, where)
# print 'item %r, count %r, minind %r' % (item, count, min_index)
return count, -min_index
# pick the highest-count/earliest item
return max(groups, key=_auxfun)[0]
然后只需添加
df['A'].fillna(most_common(df['A'].values.tolist()))
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