考虑我的数据框09-15 16:20:34.277 10119-10119/com.example D/MATCH: +19447223311
09-15 16:20:34.277 10119-10119/com.example D/MATCH: 872881122
df
如何将数据框的象限与零分开
df = pd.DataFrame(np.ones((4, 4), dtype=int), list('ABDE'), list('VWYZ'))
df
V W Y Z
A 1 1 1 1
B 1 1 1 1
D 1 1 1 1
E 1 1 1 1
答案 0 :(得分:4)
一种丑陋的方式是
In [860]: df.insert(df.shape[1]/2, 'X', 0)
In [861]: df.reindex(df.index.insert(df.shape[0]/2, 'C'), fill_value=0)
Out[861]:
V W X Y Z
A 1 1 0 1 1
B 1 1 0 1 1
C 0 0 0 0 0
D 1 1 0 1 1
E 1 1 0 1 1
,另一位来自Divarkar的灵感来自
In [897]: df.reindex(index=df.index.insert(df.shape[0]/2, 'C'),
columns=df.columns.insert(df.shape[1]/2, 'X'), fill_value=0)
Out[897]:
V W X Y Z
A 1 1 0 1 1
B 1 1 0 1 1
C 0 0 0 0 0
D 1 1 0 1 1
E 1 1 0 1 1
答案 1 :(得分:3)
df['X'] = 0
df.loc['C'] = [0]*len(df.columns)
df = df.sort_index()
df = df[[sorted(df.columns)]]
如果我无法对标签进行排序,那么:
df['X'] = 0
col = list(df.columns)
col.insert(len(col)/2,'X')
df = df[col[:-1]]
df.loc['C'] = 0
rows = list(df.index)
rows.insert(len(df)/2,'C')
df.reindex(rows[:-1])
答案 2 :(得分:2)
这是一个带有数组的数组,用np.ix_
-
a = df.values
m,n = a.shape
mask_row = np.r_[:m//2+1,m//2-1:-1:-1] < m//2
mask_col = np.r_[:n//2+1,n//2-1:-1:-1] < n//2
out = np.zeros((m+1,n+1), dtype=a.dtype)
out[np.ix_(mask_row, mask_col)] = a
df_out = pd.DataFrame(out)
df_out.columns = np.insert(df.columns, n//2, 'New')
df_out.index = np.insert(df.index, m//2, 'New')
示例输出 -
In [375]: df_out
Out[375]:
V W New Y Z
A 1 1 0 1 1
B 1 1 0 1 1
New 0 0 0 0 0
D 1 1 0 1 1
E 1 1 0 1 1
答案 3 :(得分:1)
创建新的索引和列值,然后reindex
:
a = int(len(df.columns) / 2)
cols =np.concatenate([df.columns[:a], ['X'], df.columns[a:]])
b = int(len(df.index) / 2)
idx =np.concatenate([df.index[:b], ['C'], df.index[b:]])
df = df.reindex(index=idx, columns=cols, fill_value=0)
print (df)
V W X Y Z
A 1 1 0 1 1
B 1 1 0 1 1
C 0 0 0 0 0
D 1 1 0 1 1
E 1 1 0 1 1