python3中嵌套for循环：生成两个for循环的输出_Python_Python 3.x_Loops_For Loop_Nested Loops

python3中嵌套for循环：生成两个for循环的输出

python python-3.x loops for-loop

python3中嵌套for循环：生成两个for循环的输出,python,python-3.x,loops,for-loop,nested-loops,Python,Python 3.x,Loops,For Loop,Nested Loops,我有这个密码 import numpy as np import pandas as pd from scipy.spatial import distance mn= [(252, 468), (252, 495), (274, 481), (280, 458), (298, 479), (301, 458), (324, 499)] name=['loc1','loc2','loc3','loc4','loc5','lco6','loc7'] zz= [(329, 478), (336,

我有这个密码

import numpy as np
import pandas as pd
from scipy.spatial import distance
mn= [(252, 468), (252, 495), (274, 481), (280, 458), (298, 479), (301, 458), (324, 499)]
name=['loc1','loc2','loc3','loc4','loc5','lco6','loc7']

zz= [(329, 478), (336, 455), (346, 499), (352, 478), (374, 468), (381, 499), (395, 459), (406, 488)]

L = pd.Series()
for name, i in list(zip(name, mn)):
    for e in zz:
        L[name] = distance.euclidean(e, i)
print(L)

w=100
dd = np.sqrt(np.power(L, 2) + np.power(w, 2))
print(dd)

以下是它作为L&dd输出的内容：

loc1    155.293271
loc2    154.159009
loc3    132.185476
loc4    129.522199
loc5    108.374351
lco6    109.201648
loc7     82.734515
dtype: float64
loc1    184.705170
loc2    183.752551
loc3    165.749811
loc4    163.633737
loc5    147.461859
lco6    148.070929
loc7    129.788289

我的麻烦是，它只为zz中的一个点提供了L&dd，但我希望为zz中的每个点提供L，然后能够使用它为L中的每个值获得dd

谢谢你的帮助

你就快到了。基本上，您可以用以下内容替换for循环：

L = pd.Series()
for name, i in list(zip(name, mn)):
    j = [] # save the intermediary results here
    for e in zz:
        j.append(distance.euclidean(e, i)) 
    L[name] = j # append at once all computations are done

这将为您提供如下信息：

loc1    [77.64663547121665, 85.0, 98.97979591815695, 1...
loc2    [78.85429601486528, 93.03762679690406, 94.0850...
loc3    [55.08175741568164, 67.23094525588644, 74.2159...
loc4    [52.92447448959697, 56.08029957123981, 77.6981...
loc5    [31.016124838541646, 44.94441010848846, 52.0, ...
lco6    [34.40930106817051, 35.12833614050059, 60.8769...
loc7    [21.587033144922902, 45.60701700396552, 22.0, ...

下一步，您可以使用

。应用功能：
op = (L
     .apply(lambda x: np.sqrt(np.power(x, 2) + np.power(w, 2)))
     .apply(lambda x: x[0])) # get the first value of each array

loc1    126.605687
loc2    127.349912
loc3    114.166545
loc4    113.141504
loc5    104.699570
lco6    105.754433
loc7    102.303470
dtype: float64

你就快到了。基本上，您可以用以下内容替换for循环：
L = pd.Series()
for name, i in list(zip(name, mn)):
    j = [] # save the intermediary results here
    for e in zz:
        j.append(distance.euclidean(e, i)) 
    L[name] = j # append at once all computations are done

这将为您提供如下信息：
loc1    [77.64663547121665, 85.0, 98.97979591815695, 1...
loc2    [78.85429601486528, 93.03762679690406, 94.0850...
loc3    [55.08175741568164, 67.23094525588644, 74.2159...
loc4    [52.92447448959697, 56.08029957123981, 77.6981...
loc5    [31.016124838541646, 44.94441010848846, 52.0, ...
lco6    [34.40930106817051, 35.12833614050059, 60.8769...
loc7    [21.587033144922902, 45.60701700396552, 22.0, ...

下一步，您可以使用。应用功能：
op = (L
     .apply(lambda x: np.sqrt(np.power(x, 2) + np.power(w, 2)))
     .apply(lambda x: x[0])) # get the first value of each array

loc1    126.605687
loc2    127.349912
loc3    114.166545
loc4    113.141504
loc5    104.699570
lco6    105.754433
loc7    102.303470
dtype: float64

谢谢：）对于下一个操作，我是这样做的：对于L:o.append（np.sqrt（np.power（row，2）+np.power（w，2）），dd=o它给出了[array（[126.60568708131.24404748，…]），array（[127.34991166，136.58696863，…]））那么，有办法调用dd的第一列吗？谢谢！我这样做了，但我还是一个新成员，我应该达到15才能标记它。他们注册我的标记，直到我达到15。谢谢：）对于下一个操作，我这样做了o=[]对于L:o.append（np.sqrt（np.power（row，2）+np.power（w，2）），dd=o它给出了[array]([126.60568708131.24404748，…]）、数组（[127.34991166，136.58696863，…]）那么，有办法调用dd的第一列吗？谢谢！我这样做了，但我仍然是一名新成员，我应该达到15岁才能标记。他们会注册我的标记，直到我达到15岁