示例数据如下:
unique_list = ['home0', 'page_a0', 'page_b0', 'page_a1', 'page_b1',
'page_c1', 'page_b2', 'page_a2', 'page_c2', 'page_c3']
sources = [0, 0, 1, 2, 2, 3, 3, 4, 4, 7, 6]
targets = [3, 4, 4, 3, 5, 6, 8, 7, 8, 9, 9]
values = [2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2]
使用the documentation中的示例代码
fig = go.Figure(data=[go.Sankey(
node = dict(
pad = 15,
thickness = 20,
line = dict(color = "black", width = 0.5),
label = unique_list,
color = "blue"
),
link = dict(
source = sources,
target = targets,
value = values
))])
fig.show()
这将输出以下sankey图
但是,我想在同一垂直列中以相同数字结尾的所有值,就像最左边的列的所有节点都以0结尾的方式一样。我看到了在docs中,可以移动节点的位置,但是我想知道是否有比手动输入x和y值更干净的方法来做到这一点。任何帮助表示赞赏。
答案 0 :(得分:1)
在go.Sankey()
中设置arrangement='snap'
并调整x=<list>
和y=<list>
中的x和y位置。以下设置将按要求放置节点。
图解:
请注意,在此示例中未明确设置y值。只要有一个以上的公共x值节点,y值将自动调整,以使所有节点都显示在相同的垂直位置。如果您想明确设置所有位置,只需设置arrangement='fixed'
编辑:
我添加了一个自定义函数nodify()
,该函数将相同的x位置分配给具有公共结尾的标签名称,例如'0'
中的['home0', 'page_a0', 'page_b0']
。现在,以您为例,将page_c1
更改为page_c2
,您将获得以下信息:
完整代码:
import plotly.graph_objects as go
unique_list = ['home0', 'page_a0', 'page_b0', 'page_a1', 'page_b1',
'page_c1', 'page_b2', 'page_a2', 'page_c2', 'page_c3']
sources = [0, 0, 1, 2, 2, 3, 3, 4, 4, 7, 6]
targets = [3, 4, 4, 3, 5, 6, 8, 7, 8, 9, 9]
values = [2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2]
def nodify(node_names):
node_names = unique_list
# uniqe name endings
ends = sorted(list(set([e[-1] for e in node_names])))
# intervals
steps = 1/len(ends)
# x-values for each unique name ending
# for input as node position
nodes_x = {}
xVal = 0
for e in ends:
nodes_x[str(e)] = xVal
xVal += steps
# x and y values in list form
x_values = [nodes_x[n[-1]] for n in node_names]
y_values = [0.1]*len(x_values)
return x_values, y_values
nodified = nodify(node_names=unique_list)
# plotly setup
fig = go.Figure(data=[go.Sankey(
arrangement='snap',
node = dict(
pad = 15,
thickness = 20,
line = dict(color = "black", width = 0.5),
label = unique_list,
color = "blue",
x=nodified[0],
y=nodified[1]
),
link = dict(
source = sources,
target = targets,
value = values
))])
fig.show()