Question

给出以下数据框：

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

如果我想将A列中的NaN值替换为该列（1）中重复的值，并对B列执行相同的操作，那么我需要使用哪种fillna（）？

    A       B
0   1.0     2.2
1   1.0     NaN
2   NaN     2.2

寻找通用解决方案，因为我确实拥有数千行。提前谢谢！

Answer 1

// create allocation/pointer using OpenGL GLuint vertexArray; glGenBuffers( 1,&vertexArray); glBindBuffer( GL_ARRAY_BUFFER, vertexArray); glBufferData( GL_ARRAY_BUFFER, numVertices * 16, NULL, GL_DYNAMIC_COPY ); cudaGLRegisterBufferObject( vertexArray ); void * vertexPointer; // Map the buffer to CUDA cudaGLMapBufferObject(&ptr, vertexBuffer); // Run a kernel to create/manipulate the data MakeVerticiesKernel<<<gridSz,blockSz>>>(ptr,numVerticies); // Unmap the buffer cudaGLUnmapbufferObject(vertexBuffer); // Bind the Buffer glBindBuffer( GL_ARRAY_BUFFER, vertexBuffer ); // Enable Vertex and Color arrays glEnableClientState( GL_VERTEX_ARRAY ); glEnableClientState( GL_COLOR_ARRAY ); // Set the pointers to the vertices and colors glVertexPointer(3,GL_FLOAT,16,0); glColorPointer(4,GL_UNSIGNED_BYTE,16,12); glDrawArrays(GL_POINTS,0, numVerticies); SwapBuffer();可以使用键的字典，其中键是列名。

假设您要使用最重复的值填充列，您可以使用以下方法计算字典：

fillna

Answer 2

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

Answer 3

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()))

Pandas用列值填充NaN

3 个答案: